Ch 12 - Lnk2Lrn
|CHAPTER 12 |TEMPERATURE AND HEAT |
PROBLEMS
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1. [pic] REASONING AND SOLUTION The difference between these two averages, expressed in Fahrenheit degrees, is
[pic]
Since 1 C° is equal to [pic] F°, we can make the following conversion
[pic]
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2. REASONING
a. The relationship between the Kelvin temperature T and the Celsius temperature TC is given by T = TC + 273.15 (Equation 12.1).
b. The relationship between the Kelvin temperature T and the Fahrenheit temperature TF can be obtained by following the procedure outlined in Examples 1 and 2 in the text. On the Kelvin scale the ice point is 273.15 K. Therefore, a Kelvin temperature T is T ( 273.15 kelvins above the ice point. The size of the kelvin is larger than the size of a Fahrenheit degree by a factor of [pic]. As a result a temperature that is T ( 273.15 kelvins above the ice point on the Kelvin scale is [pic](T ( 273.15) Fº above the ice point on the Fahrenheit scale. This amount must be added to the ice point of 32.0 ºF. The relationship between the Kelvin and Fahrenheit temperatures, then, is given by
[pic]
SOLUTION
a. Solving Equation 12.1 for TC, we find that
Day [pic]
Night [pic]
b. Using the equation developed in the reasoning, we find
Day [pic]
Night [pic]
3. REASONING AND SOLUTION
a. A temperature of 50.0 °F is 18.0 Fahrenheit degrees above the ice point of 32.0 °F. Since [pic], the difference of 18.0 F° is equivalent to
[pic]
Adding 10.0 Celsius degrees to the ice point of 0 °C on the Celsius scale gives a Celsius temperature of [pic]. Similar reasoning shows that a temperature of 104 °F is equivalent to a Celsius temperature of [pic].
b. The Kelvin temperature and the temperature on the Celsius scale are related by Equation 12.1: T = Tc + 273.15, where T is the Kelvin temperature and [pic] is the Celsius temperature. Therefore, a temperature of 10.0 °C is equivalent to a Kelvin temperature of 283.2 K , and a temperature of 40.0 °C is equivalent to 313.2 K .
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4. REASONING AND SOLUTION
a. The Kelvin temperature and the temperature on the Celsius scale are related by Equation 12.1: T = Tc + 273.15, where T is the Kelvin temperature and [pic] is the Celsius temperature. Therefore, a temperature of 77 K on the Celsius scale is
[pic]
b. The temperature of –196 °C is 196 Celsius degrees below the ice point of 0 °C. Since [pic], this number of Celsius degrees corresponds to
[pic]
Subtracting 353 Fahrenheit degrees from the ice point of 32.0 °F on the Fahrenheit scale gives a Fahrenheit temperature of [pic].
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5. [pic] REASONING AND SOLUTION The temperature of –273.15 °C is 273.15 Celsius degrees below the ice point of 0 °C. This number of Celsius degrees corresponds to
[pic]
Subtracting 491.67 F° from the ice point of 32.00 °F on the Fahrenheit scale gives a Fahrenheit temperature of [pic]
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6. REASONING AND SOLUTION If the voltage is proportional to the temperature difference between the junctions, then
[pic]
Thus,
[pic]
Solving for T2 yields [pic].
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7. [pic] REASONING The temperature at any pressure can be determined from the equation of the graph in Figure 12.4. Since the gas pressure is [pic] when the temperature is [pic], and the pressure is zero when the temperature is –273.15 °C, the slope of the line is given by
[pic]
Therefore, the equation of the line is [pic], or
[pic]
where [pic] when [pic].
SOLUTION Solving the equation above for T, we obtain
[pic]
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8. REASONING AND SOLUTION The space invaders temperature of 58 °I is 58 °I – 25 °Ι or 33 I° above the Space invaders ice point. Additionally, the space invaders degree is
[pic]
times smaller than the Celsius degree. Now
58 °I = (33 °I)(0.76 C°/I°) = [pic]
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9. REASONING AND SOLUTION Using the value for the coefficient of thermal expansion of steel given in Table 12.1, we find that the linear expansion of the aircraft carrier is
|ΔL = αL0 ΔT = (12 × 10–6 C(–1)(370 m)(21 °C − 2.0 °C) [pic] |(12.2) |
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10. REASONING AND SOLUTION The Concorde expands by an amount (L, where
|ΔL = αL0ΔT = (2.0 × 10–5 C°–1)(62 m)(105 °C − 23 °C)[pic] |(12.2) |
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11. [pic][pic] REASONING AND SOLUTION The steel in the bridge expands according to Equation 12.2, [pic]. Solving for [pic] and using the value for the coefficient of thermal expansion of steel given in Table 12.1, we find that the approximate length of the Golden Gate bridge is
[pic]
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12. REASONING AND SOLUTION
a. The radius of the hole will be [pic] when the plate is heated, because the hole expands as if it were made of copper.
b. The expansion of the radius is Δr = αr0ΔT. Using the value for the coefficient of thermal expansion of copper given in Table 12.1, we find that the fractional change in the radius is
Δr/r0 = αΔT = (17 × 10–6 C°–1)(110 °C − 11 °C) [pic]
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13. REASONING AND SOLUTION The value for the coefficient of thermal expansion of steel is given in Table 12.1. The relation, ΔL = α L0ΔT, written in terms of the diameter d of the rod, is
| [pic] |(12.2) |
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14. REASONING To determine the fractional decrease in length [pic], we need the decrease (L in the rod’s length. It is the sum of the decreases in the silver part and the gold part of the rod, or (L = (LSilver + (LGold. Each of the decreases can be expressed in terms of the coefficient of linear expansion (, the initial length L0, and the change in temperature (T, according to Equation 12.2.
SOLUTION Using Equation 12.2 to express the decrease in length of each part of the rod, we find the total decrease in the rod’s length to be
[pic]
The fractional decrease in the rod’s length is, then,
[pic]
Recognizing that one third of the rod is silver and two thirds is gold and taking values for the coefficients of linear expansion for silver and gold from Table 12.1, we have
[pic]
15. [pic] REASONING AND SOLUTION Assuming that the rod expands linearly with heat, we first calculate the quantity [pic] using the data given in the problem.
[pic]
Therefore, when the rod is cooled from 25.0 °C, it will shrink by
[pic]
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16. REASONING The length of either heated strip is L0 + (L, where L0 is the initial length and (L is the amount by which it expands. The expansion (L can be expressed in terms of the coefficient of linear expansion (, the initial length L0, and the change in temperature (T, according to Equation 12.2. To find the change in temperature, we will set the length of the heated steel strip equal to the length of the heated aluminum strip and solve the resulting equation for (T.
SOLUTION According to Equation 12.2, the expansion is (L = ( L0(T. Using this equation we have
[pic]
We know that the steel strip is 0.10 % longer than the aluminum strip, so that [pic]. Substituting this result into the equation above, solving for (T, and taking values for the coefficients of linear expansion for aluminum and steel from Table 12.1give
[pic]
17. REASONING AND SOLUTION ΔL = αL0ΔT gives for the expansion of the aluminum
|ΔLA = αALAΔT |(1) |
and for the expansion of the brass
|ΔLB = αBLBΔT |(2) |
Taking the coefficients of thermal expansion for aluminum and brass from Table 10.1, adding Equations (1) and (2), and solving for ΔT give
[pic]
The desired temperature is then
T = 28 °C + 21 C° = [pic]
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18. REASONING AND SOLUTION The initial diameter of the sphere, ds, is
|ds = (5.0 × 10–4)dr + dr |(1) |
where dr is the initial diameter of the ring. Applying ΔL = αL0ΔT to the diameter of the
sphere gives
|Δds = αsdsΔT |(2) |
and to the ring gives
|Δdr = αrdrΔT |(3) |
If the sphere is just to fit inside the ring, we must have
ds + Δds = dr + Δdr
Using Equations (2) and (3) in this expression and solving for ΔT give
[pic]
Substituting Equation (1) in this result and taking values for the coefficients of thermal expansion of steel and lead from Table 10.1 yield
[pic]
The final temperature is
Tf = 70.0 °C − 29 C° = [pic]
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19. [pic][pic] REASONING AND SOLUTION Recall that [pic], Equation 10.6, where [pic] is the angular frequency of the pendulum and T is the period. Using this fact and Equation 10.16, we know that the period of the pendulum before the temperature rise is given by [pic], where [pic] is the length of the pendulum. After the temperature has risen, the period becomes (using Equation 12.2), [pic]. Dividing these expressions, solving for [pic], and taking the coefficient of thermal expansion of brass from Table 12.1, we find that
[pic]
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20. REASONING Each section of concrete expands as the temperature increases by an amount (T. The amount of the expansion (L is proportional to the initial length of the section, as indicated by Equation 12.2. Thus, to find the total expansion of the three sections, we can apply this expression to the total length of concrete, which is L0 = 3(2.4 m). Since the two gaps in the drawing are identical, each must have a minimum width that is one half the total expansion.
SOLUTION Using Equation 12.2 and taking the value for the coefficient of thermal expansion for concrete from Table 12.1, we find
[pic]
The minimum necessary gap width is one half this value or
[pic]
21. REASONING AND SOLUTION
a. The ruler will try to shrink as the temperature is lowered so a [pic] is needed to keep it from shrinking.
b. The change in length of the ruler is given by ΔL = (L0ΔT, Equation 12.2. The stress needed to stretch the ruler this amount is given by Stress = F/A = Y(ΔL/L0), Equation 10.17. Substituting for ΔL and taking values for Young’s modulus Y of steel from Table 10.1 and the coefficient of thermal expansion of steel from Table 12.1, we find that
Stress = Y(ΔT = (2.0 × 1011 N/m2)(12 × 10–6 C°–1)[25 °C − (−15 °C)][pic]
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22. REASONING AND SOLUTION Let L0 = 0.50 m and L be the true length of the line at 40.0 °C. The ruler has expanded an amount
|ΔLr = L − L0 = (rL0ΔTr |(1) |
The copper plate must shrink by an amount
|ΔLp = L0 − L = (pLΔTp |(2) |
Eliminating L from Equations (1) and (2), solving for ΔTp, and using values of the coefficients of thermal expansion for copper and steel from Table 12.1, we find that
[pic]
Therefore, Tp = 40.0 °C − 14 C° = [pic].
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23. [pic] REASONING The change in length of the wire is the sum of the change in length of each of the two segments: [pic]. Using Equation 12.2 to express the changes in length, we have
[pic]
Dividing both sides by L0 and algebraically canceling ΔT gives
[pic]
The length of the steel segment of the wire is given by [pic]. Making this
substitution leads to
[pic]
This expression can be solved for the desired quantity, [pic].
SOLUTION Solving for the ratio [pic] and taking values for the coefficients of thermal expansion for aluminum and steel from Table 12.1 gives
[pic]
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24. REASONING AND SOLUTION The figure below (at the left) shows the forces that act on the middle of the aluminum wire for any value of the angle θ. The figure below (at the right) shows the same forces after they have been resolved into x and y components.
[pic]
Applying Newton's second law to the vertical forces in figure on the right gives [pic]. Solving for T gives:
|[pic] |(1) |
The following figures show how the angle is related to the initial length L0 of the wire and to the final length L after the temperature drops.
[pic]
If the distance between the supports does not change, then the distance x is the same in both figures. Thus, at the original temperature,
|x = L0 (cos θ0) |(2) |
while at the lower temperature
|x = L (cos θ) |(3) |
Equating the right hand sides of Equations (2) and (3) leads to
|[pic] |(4) |
Now, L = L0 + ΔL, and from Equation 12.2, it follows that
[pic]
Thus, Equation (4) becomes
[pic]
From the figure in the text θ0 = 3.00°. Noting that the temperature of the wire drops by 20.0 C° ((T = (20.0 C°) and taking the coefficient of thermal expansion of aluminum from Table 12.1, we find that the wire makes an angle θ with the horizontal that is
[pic]
Using this value for θ in Equation (1) gives
[pic]
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25. REASONING AND SOLUTION The volume [pic] of an object changes by an amount [pic] when its temperature changes by an amount [pic]; the mathematical relationship is given by Equation 12.3: [pic]. Thus, the volume of the kettle at 24 °C can be found by solving Equation 12.3 for [pic]. According to Table 12.1, the coefficient of volumetric expansion for copper is [pic]. Solving Equation 12.3 for [pic], we have
[pic]
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26. REASONING AND SOLUTION Taking the coefficient of volumetric expansion ( for water from Table 12.1, we find that the change in the volume of the water is
| ΔV = (V0ΔT = (207 × 10−6 C°−1)(110 m3)(27 °C − 17 °C)[pic] |(12.3) |
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27. REASONING The change (V in the interior volume of the shell is given by Equation 12.3 as (V = (V0(T, where ( is the coefficient of volume expansion, V0 is the initial volume, and (T is the increase in temperature. The interior volume behaves as if it were filled with the surrounding silver. The interior volume is spherical, and the volume of a sphere is [pic], where r is the radius of the sphere.
SOLUTION In applying Equation 12.3, we note that the initial spherical volume of the interior space is [pic], so that we have
[pic]
We have taken the coefficient of volume expansion for silver from Table 12.1.
28. REASONING The increase (V in volume is given by Equation 12.3 as (V = (V0(T, where ( is the coefficient of volume expansion, V0 is the initial volume, and (T is the increase in temperature. The lead and quartz objects experience the same change in volume. Therefore, we can use Equation 12.3 to express the two volume changes and set them equal. We will solve the resulting equation for (TQuartz.
SOLUTION Recognizing that the lead and quartz objects experience the same change in volume and expressing that change with Equation 12.3, we have
[pic]
In this result V0 is the initial volume of each object. Solving for (TQuartz and taking values for the coefficients of volume expansion for lead and quartz from Table 12.1 gives
[pic]
29. [pic] REASONING AND SOLUTION According to Equation 12.3, [pic]. Taking the coefficient of volumetric expansion ( for water from Table 12.1, we find that
[pic]
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30. REASONING AND SOLUTION Both the water and pipe expand as the temperature increases. For the expansion of the water
ΔVw = (w V0 ΔT
and for the expansion of the pipe
ΔVp = (c V0 ΔT
The initial volume of the pipe and water is V0 = (r2L The reservoir then needs a capacity of
ΔV = ΔVw − ΔVp = ((w − (c) V0 ΔT
Taking the coefficients of volumetric expansion (w and (c for water and copper from Table 12.1, we find
ΔV = (207 × 10–6 C°–1 − 51 × 10–6 C°–1)((9.5 × 10–3 m)2(76 m)(54 C°) [pic]
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31. [pic] REASONING AND SOLUTION Both the gasoline and the tank expand as the temperature increases. The coefficients of volumetric expansion (g and (s for gasoline and steel are available in Table 12.1. According to Equation 12.3, the volume expansion of the gasoline is
[pic]
while the volume of the steel tank expands by an amount
[pic]
The amount of gasoline which spills out is
[pic]
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32. REASONING AND SOLUTION Both the coffee and beaker expand as the temperature increases. For the expansion of the coffee
ΔVc = βwV0 ΔT
and for the expansion of the beaker
ΔVb = βb V0 ΔT
The excess expansion of the coffee, hence the amount which spills, is
ΔV = ΔVc − ΔVb = (βw − βb) V0 ΔT
Taking the coefficients of volumetric expansion (w and (c for coffee (water) and glass (Pyrex) from Table 12.1, we find
[pic]
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33. REASONING The change in volume of the cylindrical mercury column is proportional to the coefficient of volume expansion β for mercury, its original volume V0, and the change ΔT in temperature. The volume of a cylinder of radius r and height ΔL is [pic]. The coefficient of volume expansion for mercury can be found in Table 12.1.
SOLUTION The change in the volume of the mercury is
|[pic] |(12.3) |
Solving for ΔL gives
[pic]
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34. REASONING The change (V in volume is given by Equation 12.3 as (V = (V0(T, where ( is the coefficient of volume expansion, V0 is the initial volume, and (T is the increase in temperature. The increase in volume of the mercury is given directly by this equation, with V0 being the initial volume of the interior space of the brass shell minus the initial volume of the steel ball. If the space occupied by the mercury did not change with temperature, the spillage would simply be the increase in volume of the mercury. However, the space occupied by the mercury does change with temperature. Both the brass shell and the steel ball expand. The interior volume of the brass shell expands as if it were solid brass, and this expansion provides more space for the mercury to occupy, thereby reducing the amount of spillage. The expansion of the steel ball, in contrast, takes up space that would otherwise be occupied by mercury, thereby increasing the amount of spillage. The total spillage, therefore, is (VMercury ( (VBrass + (VSteel.
SOLUTION Table 12.1 gives the coefficients of volume expansion for mercury, brass, and steel. Applying Equation 12.3 to the mercury, the brass cavity, and the steel ball, we have
[pic]
35. [pic] REASONING In order to keep the water from expanding as its temperature increases from 15 to 25 °C, the atmospheric pressure must be increased to compress the water as it tries to expand. The magnitude of the pressure change [pic] needed to compress a substance by an amount ΔV is, according to Equation 10.20, [pic]. The ratio [pic] is, according to Equation 12.3, [pic]. Combining these two equations yields
[pic]
SOLUTION Taking the value for the coefficient of volumetric expansion ( for water from Table 12.1, we find that the change in atmospheric pressure that is required to keep the water from expanding is
[pic]
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36. REASONING AND SOLUTION
a. The apparent weight of the sphere will be larger [pic] it cools. This is because the sphere shrinks while cooling, displacing less water, and hence decreasing the buoyant force acting on it.
b. The weight of the submerged sphere before cooling is
W0 = ρgV0 − ρwgV0
The weight of the submerged sphere after cooling is
W = ρgV0 − ρwgV
The weight difference is
ΔW = W − W0 = −ρwgΔV
where ΔV = V − V0. The volume change is ΔV = βV0ΔT. Now, ΔW = −ρwgβV0ΔT. The sphere’s original volume is
V0 = (4/3)πr3 = (4/3)(π)(0.50 m)3 = 0.52 m3
The coefficient of volumetric expansion ( for aluminum is given in Table 12.1, so we have
ΔW = −(1.0 × 103 kg/m3)(9.80 m/s2)(69 × 10–6 C°–1)(0.52 m3)(−5.0 × 101 C°)[pic]
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37. [pic][pic] REASONING The cavity that contains the liquid in either Pyrex thermometer expands according to Equation 12.3, [pic]. On the other hand, the volume of mercury expands by an amount [pic], while the volume of alcohol expands by an amount [pic]. Therefore, the net change in volume for the mercury thermometer is
[pic]
while the net change in volume for the alcohol thermometer is
[pic]
In each case, this volume change is related to a movement of the liquid into a cylindrical region of the thermometer with volume [pic], where r is the radius of the region and h is the height of the region. For the mercury thermometer, therefore,
[pic]
Similarly, for the alcohol thermometer
[pic]
These two expressions can be combined to give the ratio of the heights, [pic].
SOLUTION Taking the values for the coefficients of volumetric expansion for methyl alcohol, Pyrex glass, and mercury from Table 12.1, we divide the two expressions for the heights of the liquids in the thermometers and find that
[pic]
Therefore, the degree marks are [pic] on the alcohol thermometer than on the mercury thermometer.
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38. REASONING AND SOLUTION When the temperature is 0.0 °C, P = ρ0gh0, and when the temperature is 38.0 °C, P = ρgh. Equating and solving for h gives h = (ρ0/ρ)h0. Now ρ0/ρ = V/V0, since the mass of mercury in the tube remains constant. Then h = (V/V0)h0. Now,
ΔV = V − V0 = βV0ΔT or V/V0 = 1 + βΔT
Therefore,
h = (1 + βΔT)h0 = [1 + (182 × 10−6 C°−1)(38.0 °C − 0.0 °C)](0.760 m)[pic]
where we have taken the value for the coefficient of volumetric expansion ( for mercury from Table 12.1.
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39. REASONING AND SOLUTION The heat released by the blood is given by Q = cmΔT, in which the specific heat capacity c of the blood (water) is given in Table 12.2. Then
[pic]
Therefore,
Tf = Ti − ΔT = [pic]
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40. REASONING Since the container of the glass and the liquid is being ignored and since we are assuming negligible heat exchange with the environment, the principle of conservation of energy applies. In reaching equilibrium the cooler liquid gains heat, and the hotter glass loses heat. We will apply this principle by equating the heat gained to the heat lost. The heat Q that must be supplied or removed to change the temperature of a substance of mass m by an amount (T is given by Equation 12.4 as Q = cm(T, where c is the specific heat capacity. In using this equation as we apply the energy-conservation principle, we must remember to express the change in temperature (T as the higher minus the lower temperature.
SOLUTION Applying the energy-conservation principle and using Equation 12.4 give
[pic]
Since it is the same for both the glass and the liquid, the mass m can be eliminated algebraically from this equation. Solving for cLiquid and taking the specific heat capacity for glass from Table 12.2, we find
[pic]
41. [pic] REASONING According to Equation 12.4, the heat required to warm the pool can be calculated from [pic]. The specific heat capacity c of water is given in Table 12.2. In order to use Equation 12.4, we must first determine the mass of the water in the pool. Equation 11.1 indicates that the mass can be calculated from [pic], where [pic] is the density of water and V is the volume of water in the pool.
SOLUTION Combining these two expressions, we have [pic], or
[pic]
Using the fact that [pic], the cost of using electrical energy to heat the water in the pool at a cost of $0.10 per kWh is
[pic]
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42. REASONING Since there is no heat lost or gained by the system, the heat lost by the water in cooling down must be equal to the heat gained by the thermometer in warming up. The heat Q lost or gained by a substance is given by Equation 12.4 as Q = cmΔT, where c is the specific heat capacity, m is the mass, and ΔT is the change in temperature. Thus, we have that
[pic]
We can use this equation to find the temperature of the water before the insertion of the thermometer.
SOLUTION Solving the equation above for [pic], and using the value of [pic] from Table 12.2, we have
[pic]
The temperature of the water before the insertion of the thermometer was
[pic]
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43. REASONING The metabolic processes occurring in the person’s body produce the heat that is added to the water. As a result, the temperature of the water increases. The heat Q that must be supplied to increase the temperature of a substance of mass m by an amount (T is given by Equation 12.4 as Q = cm(T, where c is the specific heat capacity. The increase (T in temperature is the final higher temperature Tf minus the initial lower temperature T0. Hence, we will solve Equation 12.4 for the desired final temperature.
SOLUTION From Equation 12.4, we have
[pic]
Solving for the final temperature, noting that the heat is Q = (3.0 × 105 J/h)(0.50 h) and taking the specific heat capacity of water from Table 12.2, we obtain
[pic]
44. REASONING AND SOLUTION We require that the heat gained by the cold water equals the heat lost by the hot water, i.e.,
Qcw = Qhw
Therefore,
ccwmcwΔTcw = chwmhwΔThw
Since the specific heat capacity is that of water in each case, ccw = chw, then
mcw(36.0 °C – 13.0 °C) = mhw(49.0 °C – 36.0 °C)
Then
mcw = (0.57)mhw
We also know that mcw + mhw = 191 kg. Substituting for mcw, and solving for mhw, we have
[pic]
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45. [pic] REASONING Let the system be comprised only of the metal forging and the oil. Then, according to the principle of energy conservation, the heat lost by the forging equals the heat gained by the oil, or [pic]. According to Equation 12.4, the heat lost by the forging is [pic], where [pic] is the final temperature of the system at thermal equilibrium. Similarly, the heat gained by the oil is given by [pic].
SOLUTION
[pic]
Solving for [pic], we have
[pic]
or
[pic]
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46. REASONING AND SOLUTION We wish to convert 2.0% of the heat Q into gravitational potential energy, i.e., (0.020)Q = mgh. Thus,
[pic]
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47. REASONING The heat Q that must be added to the water is given by Equation 12.4 as Q = cmΔT, where c is the specific heat capacity (see Table 12.2), m is the mass, and ΔT is the change in temperature. The mass of the water is related to its density ρ and volume V by m = ρV (Equation 11.1). Thus, the heat that must be added can be written as
[pic] (1)
SOLUTION Solving Equation (1) for the volume V of water and converting cubic meters to barrels, we find that
[pic]
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48. REASONING AND SOLUTION The change ΔV in volume of the water in the swimming pool is given by [pic], Equation 12.3, where β is the coefficient of volume expansion for water (see Table 12.1), V0 is the original volume of the water, and ΔT is the change in its temperature. The change in temperature depends on the heat Q added to the water by ΔT = Q/cm, Equation 12.4, where c is the specific heat capacity (see Table 12.2) and m is the mass. Substituting this expression for ΔT into the equation for ΔT gives
[pic]
But m/V0 is equal to the density ρ of water (Equation 11.1), so the change in volume of the water is
[pic]
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49. [pic] REASONING AND SOLUTION As the rock falls through a distance h, its initial potential energy [pic] is converted into kinetic energy. This kinetic energy is then converted into heat when the rock is brought to rest in the pail. If we ignore the heat absorbed by the pail, the principle of conservation of energy indicates that
[pic]
where we have used Equation 12.4 to express the heat absorbed by the rock and the water. Table 12.2 gives the specific heat capacity of the water. Solving for [pic] yields
[pic]
Substituting values yields
[pic]
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50. REASONING AND SOLUTION The heat lost by the steel rod is Q = cmΔT =
cρV0 ΔT. Table 12.2 gives the specific heat capacity c of steel. The rod contracts according to the equation: ΔL = αL0ΔT. Table 12.1 gives the coefficient of thermal expansion of steel. We also know that F = AY(ΔL/ L0), Equation 10.17, so that F = AYαΔT. Table 10.1 gives Young’s modulus Y for steel.
Combining expressions yields,
[pic]
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51. [pic] REASONING Heat Q1 must be added to raise the temperature of the aluminum in its solid phase from 130 °C to its melting point at 660 °C. According to Equation 12.4, [pic]. The specific heat c of aluminum is given in Table 12.2. Once the solid aluminum is at its melting point, additional heat Q2 must be supplied to change its phase from solid to liquid. The additional heat required to melt or liquefy the aluminum is [pic], where Lf is the latent heat of fusion of aluminum. Therefore, the total amount of heat which must be added to the aluminum in its solid phase to liquefy it is
[pic]
SOLUTION Substituting values, we obtain
[pic]
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52. REASONING AND SOLUTION
a. The latent heat of vaporization Lv of water is given in Table 12.3. To change water at 100.0 °C to steam we have from Equation 12.5 that
Q = mLv = (2.00 kg)(22.6 × 105 J/kg) = [pic]
b. For liquid water at 0.0 °C we must include the heat needed to raise the temperature to the boiling point. This heat is depends on the specific heat c of water, which is given in Table 12.2. Using Equations 12.4 and 12.5, we have Q = (cmΔT)water + mLv
Q = [4186 J/(kg(C°)](2.00 kg)(100.0 C°) + 4.52 × 106 J[pic]
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53. REASONING AND SOLUTION We want the same amount of heat removed from the water as from the ethyl alcohol, i.e.,
Qwater = Qalcohol
or
(mLf)water = (mLf)alcohol
Then, taking values of the latent heats of fusion for water and ethyl alcohol from Table 12.3, we find
[pic]
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54. REASONING As the body perspires, heat Q must be added to change the water from the liquid to the gaseous state. The amount of heat depends on the mass m of the water and the latent heat of vaporization Lv, according to Q = mLv (Equation 12.5).
SOLUTION The mass of water lost to perspiration is
[pic]
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55. [pic] REASONING From the conservation of energy, the heat lost by the mercury is equal to the heat gained by the water. As the mercury loses heat, its temperature decreases; as the water gains heat, its temperature rises to its boiling point. Any remaining heat gained by the water will then be used to vaporize the water.
According to Equation 12.4, the heat lost by the mercury is [pic]. The heat required to vaporize the water is, from Equation 12.5, [pic]. Thus, the total amount of heat gained by the water is [pic].
SOLUTION
[pic]
where [pic] and [pic]. The specific heats of mercury and water are given in Table 12.2, and the latent heat of vaporization of water is given in Table 12.3. Solving for the mass of the water that vaporizes gives
[pic]
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56. REASONING AND SOLUTION The heat required to evaporate the water is Q = mLv, and to lower the temperature of the jogger we have Q = mjcΔT. Equating these two expressions and solving for the mass of the water, m, we have
m = mjcΔT/Lv
m = (75 kg)[3500 J/(kg(C°)](1. 5 C°)/(2.42 × 106 J/kg)[pic]
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57. REASONING AND SOLUTION The heat required is Q = mLf + cmΔT, where m = (V. See Table 12.2 for the specific heat c and Table 12.3 for the latent heat Lf. Thus,
Q = (VLf + c(VΔT
Q = (917 kg/m3)(4.50 × 10–4 m)(1.25 m2){33.5 × 104 J/kg
+ [2.00 × 103 J/(kg(C°)](12.0 C°)}[pic]
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58. REASONING Since the container is being ignored and since we are assuming negligible heat exchange with the environment, the principle of conservation of energy applies in the following form: heat gained equals heat lost. In reaching equilibrium the colder aluminum gains heat in warming to 0.0 ºC, and the warmer water loses heat in cooling to 0.0 ºC. In either case, the heat Q that must be supplied or removed to change the temperature of a substance of mass m by an amount (T is given by Equation 12.4 as Q = cm(T, where c is the specific heat capacity. In using this equation as we apply the energy-conservation principle, we must remember to express the change in temperature (T as the higher minus the lower temperature. The water that freezes into ice also loses heat. The heat Q lost when a mass m of water freezes is given by Equation 12.5 as Q = mLf, where Lf is the latent heat of fusion. By including this amount of lost heat in the energy-conservation equation, we will be able to calculate the mass of water that is frozen.
SOLUTION Using the energy-conservation principle and Equations 12.4 and 12.5 gives
[pic]
Solving for mIce, taking values for the specific heat capacities from Table 12.2, and taking the latent heat for water from Table 12.3, we find that
[pic]
59. REASONING The amount Q of heat required to melt an iceberg at 0 (C is equal to mLf, where m is its mass and Lf is the latent heat of fusion for water (see Table 12.3). The mass is related to the density ρ and the volume V of the ice by Equation 11.1, m = ρ V.
SOLUTION
a. The amount of heat required to melt the iceberg is
|[pic] |(12.5) |
b. The number of years it would take to melt the iceberg is equal to the energy required to melt it divided by the energy consumed per year by the U.S.
[pic]
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60. REASONING AND SOLUTION Using the value given for the specific heat c of water given in Table 12.2, we find that the energy released by the water is
Q = (cmΔT)water + mLf
Q = [4186 J/(kg(C°)](840 kg)(10.0 C°) + (840 kg)(3.35 × 105 J/kg)[pic]
The 2.0-kW heater provides 2.0 × 103 J/s, so that the time is
[pic]
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61. [pic][pic] REASONING The system is comprised of the unknown material, the glycerin, and the aluminum calorimeter. From the principle of energy conservation, the heat gained by the unknown material is equal to the heat lost by the glycerin and the calorimeter. The heat gained by the unknown material is used to melt the material and then raise its temperature from the initial value of –25.0 °C to the final equilibrium temperature of [pic].
SOLUTION
[pic]
Taking values for the specific heat capacities of glycerin and aluminum from Table 12.2, we have
[pic]
Solving for [pic] yields,
[pic]
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62. REASONING The mass mremaining of the liquid water that remains at 100 (C is equal to the original mass m minus the mass mvaporized of the liquid water that has been vaporized. The heat Q required to vaporize this mass of liquid is given by Equation 12.5 as
Q = m Lv, where Lv is the latent heat of vaporization for water. Thus, we have
[pic]
The heat required to vaporize the water comes from the heat that is removed from the water at 0 (C when it changes phase from the liquid state to ice. This heat is also given by Equation 12.5 as Q = mLf, where Lf is the latent heat of fusion for water. Thus, the remaining mass of liquid water can be written as
[pic]
SOLUTION Using the values of Lf and Lv from Table 12.3, we find that the mass of liquid water that remains at 100 (C is
[pic]
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63. REASONING Since all of the heat generated by friction goes into the block of ice, only this heat provides the heat needed to melt some of the ice. Since the surface on which the block slides is horizontal, the gravitational potential energy does not change, and energy conservation dictates that the heat generated by friction equals the amount by which the kinetic energy decreases or [pic], where v0 and v are, respectively, the initial and final speeds and M is the mass of the block. In reality, the mass of the block decreases as the melting proceeds. However, only a very small amount of ice melts, so we may consider M to be essentially constant at its initial value. The heat Q needed to melt a mass m of water is given by Equation 12.5 as Q = mLf, where Lf is the latent heat of fusion. Thus, by equating QFriction to mLf and solving for m, we can determine the mass of ice that melts.
SOLUTION Equating QFriction to mLf and solving for m gives
[pic]
We have taken the value for the latent heat of fusion for water from Table 12.3.
64. REASONING To freeze either liquid, heat must be removed to cool the liquid to its freezing point. In either case, the heat Q that must be removed to lower the temperature of a substance of mass m by an amount (T is given by Equation 12.4 as Q = cm(T, where c is the specific heat capacity. The amount (T by which the temperature is lowered is the initial temperature T0 minus the freezing point temperature T. Once the liquid has been cooled to its freezing point, additional heat must be removed to convert the liquid into a solid at the freezing point. The heat Q that must be removed to freeze a mass m of liquid into a solid is given by Equation 12.5 as Q = mLf, where Lf is the latent heat of fusion. The total heat to be removed, then, is the sum of that specified by Equation 12.4 and that specified by Equation 12.5, or QTotal = cm (T0 ( T) + mLf. Since we know that same amount of heat is removed from each liquid, we can set QTotal for liquid A equal to QTotal for liquid B and solve the resulting equation for Lf, A ( Lf, B.
SOLUTION Setting QTotal for liquid A equal to QTotal for liquid B gives
cAm (T0 ( TA) + mLf, A = cBm (T0 ( TB) + mLf, B
Noting that the mass m can be eliminated algebraically from this result and solving for Lf, A ( Lf, B, we find
[pic]
65. [pic] REASONING In order to melt, the bullet must first heat up to 327.3 °C (its melting point) and then undergo a phase change. According to Equation 12.4, the amount of heat necessary to raise the temperature of the bullet to 327.3 °C is [pic], where m is the mass of the bullet. The amount of heat required to melt the bullet is given by [pic], where [pic] is the latent heat of fusion of lead.
The lead bullet melts completely when it comes to a sudden halt; all of the kinetic energy of the bullet is converted into heat; therefore,
[pic]
The value for the specific heat c of lead is given in Table 12.2, and the value for the latent heat of fusion Lf of lead is given in Table 12.3. This expression can be solved for v, the minimum speed of the bullet for such an event to occur.
SOLUTION Solving for v, we find that the minimum speed of the lead bullet is
[pic]
[pic]
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66. REASONING AND SOLUTION The steel band must be heated so that it can expand to fit the wheel. The diameter of the band must increase in length by an amount[pic]. Also, ΔL = αL0ΔT, where the coefficient of thermal expansion ( of steel is given in Table 12.1, so that
ΔT = ΔL/α L0 = (6.00 × 10–4 m)/[(12 × 10–6/C°)(1.00 m)] = 5.0 × 101 C°
The heat to expand the steel band comes from the heat released from the steam as it changes to water. Therefore Qstm = Qsb, or
mstmLv + (cmstmΔT)water = (cmsbΔT)
Substituting the values for the latent heat of vaporization of water (see Table 12.3), the specific heat capacity of water (see Table 12.2), and the specific heat capacity of steel (see Table 12.2) gives
mstm(22.6 × 105 J/kg) + [4186 J/(kg(C°)]mstm(100.0 °C − 70.0 °C)
= [452 J/(kg(C°)](25.0 kg)(50.0 C°)
Solving for the mass of the steam, we obtain[pic].
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67. [pic] REASONING AND SOLUTION From inspection of the graph that accompanies this problem, a pressure of [pic] corresponds to a temperature of 0 °C. Therefore, liquid carbon dioxide exists in equilibrium with its vapor phase at [pic] when the vapor pressure is [pic].
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68. REASONING The definition of percent relative humidity is given by Equation 12.6 as follows:
[pic]
Using R to denote the percent relative humidity, P to denote the partial pressure of water vapor, and PV to denote the equilibrium vapor pressure of water at the existing temperature, we can write Equation 12.6 as
[pic]
The partial pressure of water vapor P is the same at the two given temperatures. The relative humidity is not the same at the two temperatures, however, because the equilibrium vapor pressure PV is different at each temperature, with values that are available from the vapor pressure curve given with the problem statement. To determine the ratio R10/R40, we will apply Equation 12.6 at each temperature.
SOLUTION Using Equation 12.6 and reading the values of R10 and R40 from the vapor pressure curve given with the problem statement, we find
[pic]
69. [pic] REASONING AND SOLUTION From the vapor pressure curve that accompanies Problem 68, it is seen that the partial pressure of water vapor in the atmosphere at 10 °C is about 1400 Pa, and that the equilibrium vapor pressure at 30 °C is about 4200 Pa. The relative humidity is, from Equation 12.6,
[pic][pic]
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70. REASONING To bring the water to the point where it just begins to boil, its temperature must be increased to the boiling point. The heat Q that must be added to raise the temperature of a substance of mass m by an amount (T is given by Equation 12.4 as Q = cm(T, where c is the specific heat capacity. The amount (T by which the temperature changes is the boiling temperature minus the initial temperature of 100.0 ºC. The boiling temperature is the temperature at which the vapor pressure of the water equals the external pressure of 3.0 × 105 Pa and can be read from the vapor pressure curve for water given in Figure 12.32.
SOLUTION Using Equation 12.4, with TBP being the boiling temperature and T0 being the initial temperature, we have
[pic]
According to Figure 12.32, an external pressure of 3.0 × 105 Pa corresponds to a boiling point temperature of TBP = 134 ºC. Using this value in Equation 12.4 and taking the specific heat capacity for water from Table 12.2, we determine the heat to be
[pic]
71. REASONING AND SOLUTION At a temperature of 30.0 °C the equilibrium vapor pressure is 4200 Pa (as seen from the vapor pressure curve for water that accompanies Problem 68).
| [pic] |(12.6) |
The partial pressure is equal to
[pic]
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72. REASONING AND SOLUTION The pressure inside the container is due to the weight on the piston in addition to the pressure of the atmosphere. The 120-kg mass produces a pressure of
[pic]
If we include atmospheric pressure (1.01 × 105 Pa), the total pressure inside the container is
Ptotal = P + Patm = 2.0 × 105 Pa
Examination of the vaporization curve for water in Figure 12.32 shows that the temperature corresponding to this pressure at equilibrium is[pic].
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73. [pic] REASONING We must first find the equilibrium temperature [pic] of the iced tea. Once this is known, we can use the vapor pressure curve that accompanies Problem 68 to find the partial pressure of water vapor at that temperature and then estimate the relative humidity using Equation 12.6.
According to the principle of energy conservation, when the ice is mixed with the tea, the heat lost by the tea is gained by the ice, or [pic]. The heat gained by the ice is used to melt the ice at 0.0 °C; the remainder of the heat is used to bring the water at 0.0 °C up to the final equilibrium temperature [pic].
SOLUTION
[pic]
The specific heat capacity of water is given in Table 12.2, and the latent heat of fusion Lf of water is given in Table 12.3. Solving for [pic], we have
[pic]
According to the vapor pressure curve that accompanies Problem 68, at a temperature of
9.91 °C, the equilibrium vapor pressure is approximately 1250 Pa. At 30 °C, the equilibrium vapor pressure is approximately 4400 Pa. Therefore, according to Equation 12.6, the percent relative humidity is approximately
[pic]
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74. REASONING AND SOLUTION Equation 12.6 defines the relative humidity as
[pic]
According to the vapor pressure curve that accompanies Problem 68, the equilibrium vapor pressure of water at 36°C is 5.8 x 103 Pa. Since the water condenses on the coils when the temperature of the coils is 30 °C, the relative humidity at 30 °C is 100 percent. From Equation 12.6 this implies that the partial pressure of water vapor in the air must be equal to the equilibrium vapor pressure of water at 30 °C. From the vapor pressure curve, this pressure is 4.4 x 103 Pa. Thus, the relative humidity in the room is
[pic]
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75. REASONING AND SOLUTION At 10 °C the equilibrium vapor pressure is 1250 Pa. At 25 °C, the equilibrium vapor pressure is 3200 Pa. To get the smallest possible value for the relative humidity assume that the air at 10 °C is saturated. That is, take the partial pressure to be 1250 Pa (see the vapor pressure curve for water that accompanies Problem 68). Then we have,
[pic]
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76. REASONING AND SOLUTION The water will boil if the vapor pressure of the water is equal to the ambient pressure. The pressure at a depth h in the water can be determined from Equation 11.4: [pic]. When h = 10.3 m,
[pic]
The vapor pressure curve in Figure 12.32 shows that the vapor pressure of water is equal to[pic] at a temperature of 123 °C. Thus, the water at that depth has a temperature of [pic].
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77. REASONING The change in length ΔL of the pipe is proportional to the coefficient of linear expansion α for steel, the original length L0 of the pipe, and the change in temperature ΔT . The coefficient of linear expansion for steel can be found in Table 12.1.
SOLUTION The change in length of the pipe is
|[pic] |(12.2) |
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78. REASONING AND SOLUTION The block of ice must undergo a change in temperature, followed by a change in phase, and then another change in temperature,
Q = Qice + Qice-to-water + Qwater
Q = (cmΔT)ice + mLf + (cmΔT)water
Table 12.2 gives values for the specific heats of ice and water. Table 12.3 gives the latent heat of fusion of water. Using these values, we have
4.11 × 106 J = [2.00 × 103 J/(kg(C°)](10.0 kg)(10.0 C°) + (10.0 kg)(33.5 × 104 J/kg)
+ [4186 J/(kg(C°)](10.0 kg)Tf
Solving for Tf we obtain, [pic].
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79. [pic] REASONING From the principle of conservation of energy, the heat lost by the coin must be equal to the heat gained by the liquid nitrogen. The heat lost by the silver coin is, from Equation 12.4, [pic] (see Table 12.2 for the specific heat capacity of silver). If the liquid nitrogen is at its boiling point, –195.8 °C, then the heat gained by the nitrogen will cause it to change phase from a liquid to a vapor. The heat gained by the liquid nitrogen is [pic], where [pic] is the mass of liquid nitrogen that vaporizes, and [pic] is the latent heat of vaporization for nitrogen (see Table 12.3).
SOLUTION
[pic]
Solving for the mass of the nitrogen that vaporizes
[pic]
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80. REASONING Since there is no heat lost or gained by the system, the heat lost by the coffee in cooling down must be equal to the heat gained by the ice as it melts plus the heat gained by the melted water as it subsequently heats up. The heat Q lost or gained by a substance is given by Equation 12.4 as Q = cmΔT, where c is the specific heat capacity (see Table 12.2), m is the mass, and ΔT is the change in temperature. The heat that is required to change ice at 0 (C into liquid water at 0 (C is given by Equation 12.5 as Q = miceLf, where mice is the mass of ice and Lf is the latent heat of fusion for water (see Table 12.3). Thus, we have that
[pic]
The mass of the coffee can be expressed in terms of its density as mcoffee= ρcoffeeVcoffee (Equation 11.1). The change in temperature of the coffee is ΔTcoffee= 85 (C – T, where T is the final temperature of the coffee. The change in temperature of the water is ΔTwater =
T – 0 (C. With these substitutions, the equation above becomes
[pic]
Solving this equation for the final temperature gives
[pic]
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81. REASONING AND SOLUTION The fractional change in the length of the beam is given by ΔL = αL0ΔT to be ΔL/L0 = αΔT. Taking the value for the coefficient of thermal expansion ( for steel from Table 12.1, we find that
[pic]
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82. REASONING AND SOLUTION The change in the coin’s diameter is Δd = α d0ΔT, according to Equation 12.2. Solving for ( gives
|[pic] |(12.2) |
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83. [pic] REASONING AND SOLUTION The cider will expand according to Equation 12.3; therefore, the change in volume of the cider is
[pic]
At a cost of two dollars per gallon, this amounts to
[pic] or [pic]
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84. REASONING We assume that no heat is lost through the chest to the outside. Then, energy conservation dictates that the heat gained by the soda is equal to the heat lost by the watermelon in reaching the final temperature [pic]. Each quantity of heat is given by Equation 12.4, [pic], where we write the change in temperature [pic] as the higher temperature minus the lower temperature.
SOLUTION Starting with the statement of energy conservation, we have
[pic]
[pic]
Since the watermelon is being treated like water, we take the specific heat capacity of water from Table 12.2. Thus, the above equation becomes
[pic]
Suppressing units for convenience and algebraically simplifying, we have
[pic]
Solving for [pic], we obtain
[pic]
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85. REASONING AND SOLUTION
[pic]
Using the vaporization curve from Problem 68, we see that at 27 °C, the equilibrium vapor pressure is 3800 Pa. The partial pressure is, therefore,
[pic]
From the vaporization curve accompanying Problem 68 we see that the dew point (100% relative humidity at equilibrium temperature) is
[pic]
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86. REASONING AND SOLUTION The Rankine and Fahrenheit degrees are the same size, since the difference between the steam point and ice point temperatures is the same for both. The difference in the ice points of the two scales is 491.67 – 32.00 = 459.67. To get Rankine from Fahrenheit this amount must be added, so [pic]
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87. [pic] REASONING According to the statement of the problem, the initial state of the system is comprised of the ice and the steam. From the principle of energy conservation, the heat lost by the steam equals the heat gained by the ice, or [pic]. When the ice and the steam are brought together, the steam immediately begins losing heat to the ice. An amount [pic] is released as the temperature of the steam drops from 130 °C to 100 °C, the boiling point of water. Then an amount of heat [pic] is released as the steam condenses into liquid water at 100 °C. The remainder of the heat lost by the "steam" [pic] is the heat that is released as the water at 100 °C cools to the equilibrium temperature of [pic]. According to Equation 12.4, [pic] and [pic] are given by
[pic] and [pic]
[pic] is given by [pic], where [pic] is the latent heat of vaporization of water. The total heat lost by the steam has three effects on the ice. First, a portion of this heat [pic] is used to raise the temperature of the ice to its melting point at 0.00 °C. Then, an amount of heat [pic] is used to melt the ice completely (we know this because the problem states that after thermal equilibrium is reached the liquid phase is present at 50.0 °C). The remainder of the heat [pic] gained by the "ice" is used to raise the temperature of the resulting liquid at 0.0 °C to the final equilibrium temperature. According to Equation 12.4, [pic] and [pic] are given by
[pic] and [pic]
[pic] is given by [pic], where [pic] is the latent heat of fusion of ice.
SOLUTION According to the principle of energy conservation, we have
[pic]
or
[pic]
[pic]
Values for specific heats are given in Table 12.2, and values for the latent heats are given in Table 12.3. Solving for the ratio of the masses gives
[pic]
[pic]
or
[pic]
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88. REASONING AND SOLUTION
a. According to Equation 12.5 and the definition of the density ( given in Equation 11.1, it follows that Q = mLv = (VLv. Taking the latent heat of vaporization of water from Table 12.3 and the density of water from Table 11.1, we find that
Q= (1.000 × 103 kg/m3)(2.59 × 106 m2)(0.0254 m)(22.6 × 105 J/kg)[pic]
b. [pic]
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89. REASONING According to Equation 6.10b, the average power is the change in energy divided by the time. The change in energy in this problem is the heat supplied to the water and the coffee mug to raise their temperature from 15 to 100 ºC, which is the boiling point of water. The time is given as three minutes (180 s). The heat Q that must be added to raise the temperature of a substance of mass m by an amount (T is given by Equation 12.4 as Q = cm(T, where c is the specific heat capacity. This equation will be used for the water and the material of which the mug is made.
SOLUTION Using Equation 6.10b, we write the average power [pic] as
[pic]
The heats QWater and QMug each can be expressed with the aid of Equation 12.4, so that we obtain
[pic]
The specific heat of water has been taken from Table 12.2.
90. REASONING The force Fapplied to compress a spring by an amount ΔL is given by Equation 10.1 as Fapplied = kΔL, where k is the spring constant of the spring. The change in length of the spring is related to its initial length L0 and the change in temperature ΔT by Equation 12.2, ΔL = αL0ΔT, where α is the coefficient of linear expansion for brass. Thus, the applied force is
[pic]
SOLUTION Using the value of α for brass from Table 12.1, we have
[pic]
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91. [pic] REASONING AND SOLUTION According to Equation 12.4, the total heat per kilogram required to raise the temperature of the water is
[pic]
The mass flow rate, [pic], is given by Equations 11.7 and 11.10 as [pic]=[pic], where [pic] and [pic] are the density and volume flow rate, respectively. We have, therefore,
[pic]
Therefore, the minimum power rating of the heater must be
[pic]
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92. REASONING AND SOLUTION
a. As the wheel heats up, it will expand. Its radius, and therefore, its moment of inertia, will increase. Since no net external torque acts on the wheel, conservation of angular momentum applies where, according to Equation 9.10, the angular momentum is given by: [pic], where I is the moment of inertia and ω is the angular velocity. When the moment of inertia increases at the higher temperature, the angular speed must decrease in order for the angular momentum to remain the same.
Thus, the angular speed of the wheel decreases as the wheel heats up .
b. According to the principle of conservation of angular momentum,
[pic]
Solving for [pic], we have, treating the bicycle wheel as a thin-walled hollow hoop ([pic], see Table 9.1)
[pic]
According to Equation 12.2, [pic], and the final radius of the wheel at the higher temperature is,
[pic]
Therefore, taking the coefficient of thermal expansion ( for steel from Table 12.1, we find that the angular speed of the wheel at the higher temperature is
[pic]
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93. REASONING AND SOLUTION The steel ball and aluminum plate expand when heat is added to the system. Once the correct amount of heat has been added, the two have the same radius and the ball fits into the plate. Upon heating, the change in length is given by ΔL = αL0ΔT, where ΔL = (Lf − L0).
We require that the final lengths of the steel and aluminum be equal, i.e., (Lf)st = (Lf)al. Substituting for ΔL and equating the final lengths yields
(Lf)st = (L0)st(1 + αstΔT) = (Lf)al = (L0)al(1 + αalΔT)
Solving for ΔT and using values for the coefficients of thermal expansion for aluminum and steel from Table 12.1, we find that
ΔT = (0.0010)/(αal − 1.0010 αst) = 91 C°
Taking the values for the specific heat capacities for steel and aluminum from Table 12.2, we find that the amount of heat that must be added is, therefore,
Q = (cmΔT)st + (cmΔT)al
Q = {[452 J/(kg(C°)](1.5 kg) + [9.00 × 102 J/(kg(C°)](0.85 kg)}(91 C°)[pic]
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94. REASONING AND SOLUTION Let L0 be the length of the wire before heating. After heating the wire will have stretched an amount ΔLw = αw L0ΔT, while the gap in the concrete "stretches" an amount ΔLc = αc L0ΔT.
The net change in length of the wire is then ΔLc − ΔLw = (αc − αw) L0ΔT. The additional stress needed to produce this change is
Stress = Y(ΔLc − ΔLw)/ L0 = Y (αc − αw)ΔT
Taking values for the coefficients of thermal expansion of concrete and steel from Table 12.1 and the value for Young’s modulus Y of aluminum from Table 10.1, we find that the additional tension in the wire is
[pic]
The new tension in the wire is, then, 50.0 N – 32 N = [pic]
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95. CONCEPT QUESTIONS
a. 1 A( is larger than 1 B(, because there are only 90 A( between the ice and boiling points of water, while there are 110 B( between these points.
b. +20 (A is hotter than +20 (B, because +20 (A is 50 A( above the ice point of water, while +20 (B is at the ice point.
SOLUTION
a. Since there are 90.0 A( and 110.0 B( between the ice and boiling points of water, we have that
[pic]
b. +40.0 (A is 70.0 A( above the ice point of water. On the B thermometer, this is
[pic] above the ice point.
The temperature on the B scale is
[pic]
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96. CONCEPT QUESTIONS
a. According to Equation 12.2, the factors that determine the amount ΔL by which the length of a rod changes are the coefficient of linear expansion (, its initial length L0, and the change in temperature ΔT.
b. Since the materials from which the rods are made are different, the coefficients of linear expansion are different. The change in length is the same for each rod when the change in temperature is the same, however. Therefore, the initial lengths must be different to compensate for the fact that the expansion coefficients are different.
SOLUTION The change in length of the lead rod is (from Equation 12.2)
|[pic] |(1) |
Similarly, the change in length of the quartz rod is
|[pic] |(2) |
where the temperature change ΔT is the same for both. Since both rods change length by the same amount, ΔLL = ΔLQ. Equating Equations (1) and (2) and solving for LQ yields
[pic]
Values for the coefficients of thermal expansion for lead and quartz have been taken from Table 12.1.
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97. CONCEPT QUESTIONS
a. According to Equation 12.3, the change ΔV in volume depends on the coefficient of volume expansion β, the initial volume V0, and the change in temperature ΔT.
b. The liquid expands more, because its coefficient of volume expansion is larger and the change in volume is directly proportional to that coefficient.
c. The volume of liquid that spills over is equal to the change in volume of the liquid minus the change in volume of the can.
SOLUTION The volume of liquid that spills over the can is the difference between the increase in the volume of the liquid and that of the aluminum can:
[pic]
Therefore,
[pic]
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98. CONCEPT QUESTIONS
a. The change in temperature is determined by the amount Q of heat added, the specific heat capacity c and mass m of the material (see Equation 12.4).
b. The heat and the mass are the same, but the changes in temperature are different. The only factor that can account for the different temperature changes is the specific heat capacities, which must be different.
SOLUTION The identity of the second bar can be made by determining its specific heat capacity and making a comparison with the values in Table 12.2. The heat supplied to each bar is given by Equation 12.4: [pic]. Since identical amounts of heat are supplied to each bar, we have
[pic] or [pic]
Assigning c1 to glass, we have c1 = 840 J/(kg(C() from Table 12.2. Solving for c2, we have
[pic]
Inspection of Table 12.2 shows that the second bar is made of [pic].
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99. CONCEPT QUESTIONS
a. The amount of heat required to melt an object is directly proportional to the latent heat of fusion, according to Equation 12.5. Since each has the same mass and more heat is required to melt B, it has the larger latent heat of fusion.
b. The amount of heat required to melt an object is directly proportional to its mass, according to Equation 12.5. If the mass is doubled, the heat required to melt the object also doubles.
SOLUTION
a. According to Equation 12.5, the latent heats of fusion for A and B are
[pic]
b. The amount of heat required to melt object A when its mass is 6.0 kg is
|[pic] |(12.5) |
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100. CONCEPT QUESTIONS
a. It does not mean that the partial pressure of water vapor in the air equals atmospheric pressure. The partial pressure is less than atmospheric pressure.
b. No, the humidity is not 100%. The vapor pressure at the higher temperature is greater than that at the lower temperature, but the partial pressure of the water vapor in the air has remained the same. According to Equation 12.6, the humidity has fallen below 100%.
SOLUTION
a. The percentage of atmospheric pressure is
[pic]
b. The percentage is [pic].
c. The relative humidity at 35 (C is
|[pic] |(12.6) |
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101. CONCEPT QUESTIONS
a. When the ball and the plate are both heated to a higher common temperature, the ball passes through the hole. Since the ball’s diameter is greater than the hole’s diameter to start with, this must mean that the hole expands more than the ball for the same temperature change. The hole expands as if it were filled with the material that surrounds it. We conclude, therefore, that the coefficient of linear expansion for the plate is greater than that for the ball.
b. In each arrangement, the ball’s diameter exceeds the hole’s diameter by the same amount, the diameters of the holes are the same, the diameters of the balls are the same, and the initial temperatures are the same. The only difference between the various arrangements, then, is in the coefficients of linear expansion. The ball and the hole are both expanding. However, the hole is expanding more than the ball, to the extent that the coefficient of linear expansion of the material of the plate exceeds that of the ball. Thus, we need to examine the difference between the two coefficients, in order to decide the order in which the balls fall through the holes as the temperature increases. Referring to Table 12.1, we find the following differences for each of the arrangements:
|Arrangement I |(Lead ( (Gold = 29 × 10(6 (Cº)(1 ( 14 × 10(6 (Cº)(1 = 15 × 10(6 (Cº)(1 |
|Arrangement II |(Aluminum ( (Steel = 23 × 10(6 (Cº)(1 ( 12 × 10(6 (Cº)(1 = 11 × 10(6 (Cº)(1 |
|Arrangement III |(Silver ( (Quartz = 19 × 10(6 (Cº)(1 ( 0.50 × 10(6 (Cº)(1 = 18.5 × 10(6 (Cº)(1 |
In Arrangement III the coefficient of linear expansion of the plate-material exceeds that of the ball-material by the greatest amount. Therefore, the quartz ball will fall through the hole first. Next in sequence is Arrangement I, so the gold ball will fall second. Last is Arrangement II, so the steel ball will be the last to fall.
SOLUTION According to Equation 12.2, the diameter increases by an amount (D = (D0(T when the temperature increases by an amount (T, where D0 is the initial diameter and ( is the coefficient of linear expansion. Thus, we can write the final diameter as D = D0 + (D0(T. Since the diameters of the ball and the hole are the same when the ball falls through the hole, we have
[pic]
Solving for the change in temperature, we obtain
[pic]
[pic]
Since each arrangement has an initial temperature of 25.0 ºC, the temperatures at which the balls fall through the holes are as follows:
Arrangement I T = 25.0 ºC + 6.7 Cº = 31.7 ºC
Arrangement II T = 25.0 ºC + 9.1 Cº = 34.1 ºC
Arrangement III T = 25.0 ºC + 5.4 Cº = 30.4 ºC
These results are consistent with our answer to Concept Question (b).
102. CONCEPT QUESTIONS
a. Two portions of the same liquid that have the same mass, but different initial temperatures, when mixed together, will yield an equilibrium mixture that has a temperature lying exactly midway between the two initial temperatures. That is, the final temperature is [pic]. This occurs only when the mixing occurs without any exchange of heat with the surroundings. Then, all of the heat lost from the warmer portion is gained by the cooler portion. Since each portion is identical except for temperature, the warmer portion cools down by the same number of degrees that the cooler portion warms up, yielding a mixture whose final equilibrium temperature is midway between the two initial temperatures.
b. There are two ways to apply the logic described in Concept Question (a). Portions A and B have the same mass m, so they will yield a combined mass of 2m. Thus, we can imagine portions A and B mixed together to yield an equilibrium temperature of [pic]. This mixture has a mass 2m, just like the mass of portion C, so when it is mixed with portion C, the final equilibrium temperature will be [pic]. The other way to apply the logic from Concept Question (a) is to mix half of portion C with portion A and half with portion B. This will produce two mixtures with different temperatures and each having a mass 2m, which can then be combined. The final equilibrium temperature of this combination is again 60.0 ºC.
SOLUTION Since we are assuming negligible heat exchange with the surroundings, the principle of conservation of energy applies in the following form: heat lost equals heat gained. In reaching equilibrium the warmer portions lose heat and the cooler portions gain heat. In either case, the heat Q that must be supplied or removed to change the temperature of a substance of mass m by an amount (T is given by Equation 12.4 as Q = cm(T, where c is the specific heat capacity. We will assume that the final temperature is between 94.0 ºC and 78.0 ºC. We could also assume that the final temperature is between 78.0 ºC and 34.0 ºC. With either assumption we would obtain the same answer for the final temperature, provided that we remember to express the change in temperature (T as the higher minus the lower temperature in applying the energy-conservation principle. We have, then,
[pic]
The specific heat capacities for each portion have the same value c, while mA = mB = m and mC = 2m. With these substitutions, we find
[pic]
This answer agrees with our answer to Concept Question (b).
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