Chapter 18 Heat and the First Law of Thermodynamics

[Pages:94]Chapter 18 Heat and the First Law of Thermodynamics

Conceptual Problems

1 ? Object A has a mass that is twice the mass of object B and object A

has a specific heat that is twice the specific heat of object B. If equal amounts of

heat are transferred to these objects, how do the subsequent changes in their

temperatures compare? (a) TA = 4TB , (b) TA = 2TB , (c) TA = TB ,

(d) TA

=

1 2

TB

,

(e)

TA

=

1 4

TB

Picture the Problem We can use the relationship Q = mcT to relate the temperature changes of objects A and B to their masses, specific heats, and the amount of heat supplied to each.

Express the change in temperature of object A in terms of its mass, specific heat, and the amount of heat supplied to it:

TA

=

Q mAcA

Express the change in temperature of object B in terms of its mass, specific heat, and the amount of heat supplied to it:

Divide the second of these equations by the first to obtain:

Substitute and simplify to obtain:

TB

=

Q mBcB

TB = mAcA TA mBcB

TB TA

=

(2mB )(2cB )

mBcB

= 4 TA

=

1 4

TB

and (e) is correct.

2 ? Object A has a mass that is twice the mass of object B. The temperature change of object A is equal to the temperature change of object B when the objects absorb equal amounts of heat. It follows that their specific heats are related by (a) cA = 2cB , (b) 2cA = cB , (c) cA = cB , (d) none of the above

Picture the Problem We can use the relationship Q = mcT to relate the temperature changes of objects A and B to their masses, specific heats, and the amount of heat supplied to each.

1685

1686 Chapter 18

Relate the temperature change of object A to its specific heat and mass:

Relate the temperature change of object B to its specific heat and mass:

Equate the temperature changes and simplify to obtain:

Solve for cA:

TA

=

Q mAcA

TB

=

Q mBcB

1= 1 mBcB mAcA

cA

=

mB mA

cB

=

mB 2mB

cB

=

1 2

cB

and (b) is correct.

3 ? [SSM] The specific heat of aluminum is more than twice the specific heat of copper. A block of copper and a block of aluminum have the same mass and temperature (20?C). The blocks are simultaneously dropped into a single calorimeter containing water at 40?C. Which statement is true when thermal equilibrium is reached? (a) The aluminum block is at a higher temperature than the copper block. (b) The aluminum block has absorbed less energy than the copper block. (c) The aluminum block has absorbed more energy than the copper block. (d) Both (a) and (c) are correct statements.

Picture the Problem We can use the relationship Q = mcT to relate the amount

of energy absorbed by the aluminum and copper blocks to their masses, specific heats, and temperature changes.

Express the energy absorbed by the aluminum block:

QAl = mAlcAlT

Express the energy absorbed by the copper block:

QCu = mCucCuT

Divide the second of these equations by the first to obtain:

Because the block's masses are the same and they experience the same change in temperature:

QCu = mCucCuT QAl mAlcAlT

QCu = cCu < 1 QAl cAl or QCu < QAl and

(c) is correct.

Heat and the First Law of Thermodynamics 1687

4 ? A block of copper is in a pot of boiling water and has a temperature of 100?C. The block is removed from the boiling water and immediately placed in an insulated container filled with a quantity of water that has a temperature of 20?C and has the same mass as the block of copper. (The heat capacity of the insulated container is negligible.) The final temperature will be closest to (a) 40?C, (b) 60?C, (c) 80?C.

Determine the Concept We can use the relationship Q = mcT to relate the temperature changes of the block of copper and the boiling water to their masses, specific heats, and the amount of heat supplied to or absorbed by each.

Relate the heat supplied by the block of copper to its specific heat, mass, and temperature change as it cools to its equilibrium temperature:

QCu = mCucCuTCu

( ) = mCucCu Tequil -100?C

Relate the heat absorbed by the water to its specific heat, mass, and temperature change as it warms to its equilibrium temperature:

QH2O = mH2OcH2OTH2O

( ) = m c H2O H2O Tequil - 20?C

Because thermal energy is conserved in this mixing process:

Q = QCu + QH2O = 0 or

( ) ( ) mCucCu Tequil -100?C + m c H2O H2O Tequil - 20?C = 0

Because the mass of the water is equal to the mass of the block of copper:

( ) ( ) cCu Tequil -100?C + cH2O Tequil - 20?C = 0

Solving for Tequil yields:

( ) ( ) Tequil =

100?C cCu + 20?C cH2O cCu + cH2O

Substitute numerical values (See Table 18-1 for the specific heats of copper and water.)and evaluate Tequil:

Tequil

=

(100?C)

0.386

kJ kg C?

+

(20?C)

4.18

0.386 kJ + 4.18 kJ

kJ kg C?

27?C

(a)

is correct.

kg C?

kg C?

1688 Chapter 18

5 ? You pour both a certain amount of water at 100?C and an equal amount of water at 20?C into an insulated container. The final temperature of the mixture will be (a) 60?C, (b) less than 60?C, (c) greater than 60?C.

Determine the Concept We can use the relationship Q = mcT to relate the temperature changes of the room-temperature water and the boiling water to their masses, specific heats, and the amount of heat supplied to or absorbed by each.

Because thermal energy is conserved in this mixing process:

Q = Qhot H2O + QH2O = 0 or

( ) ( ) m c hot H2O hot H2O Tequil -100?C + mH2OcH2O Tequil - 20?C = 0

Because the mass of the boiling water is equal to the mass of the water that is initially at 20?C and the specific heat of water is independent of its temperature:

Tequil -100?C + Tequil - 20?C = 0

Solving for Tequil yields:

Tequil = 60?C

and (a) is correct.

6 ? You pour some water at 100?C and some ice cubes at 0?C into an insulated container. The final temperature of the mixture will be (a) 50?C, (b) less than 50?C but larger than 0?C, (c) 0?C, (d) You cannot tell the final temperature from the data given.

Determine the Concept We would need to know the mass of the boiling water and the mass of the melting ice in order to determine the final temperature of the

mixture. (d ) is correct.

7 ? You pour water at 100oC and some ice cubes at 0?C into an insulated container. When thermal equilibrium is reached, you notice some ice remains and floats in liquid water. The final temperature of the mixture is (a) above 0?C, (b) less than 0?C, (c) 0?C, (d) You cannot tell the final temperature from the data given

Determine the Concept Because some ice remains when thermal equilibrium is reached, the equilibrium temperature must be the temperature of the ice cubes at

their melting point. (c) is correct.

Heat and the First Law of Thermodynamics 1689

8 ? Joule's experiment establishing the mechanical equivalence of heat involved the conversion of mechanical energy into internal energy. Give some everyday examples in which some of the internal energy of a system is converted into mechanical energy.

Determine the Concept One example from everyday life is that of a hot gas that expands and does work-as in the piston of an engine. Another example is the gas escaping from a can of spray paint. As it escapes, it moves air molecules and, in the process, does work against atmospheric pressure.

9 ? Can a gas absorb heat while its internal energy does not change? If so, give an example. If not, explain why not.

Determine the Concept Yes. Eint = Qin + Won . If the gas does work at the same rate that it absorbs heat, its internal energy will remain constant.

10 ? The equation Eint = Q + W is the formal statement of the first law of thermodynamics. In this equation, the quantities Q and W, respectively, represent (a) the heat absorbed by the system and the work done by the system, (b) the heat absorbed by the system and the work done on the system, (c) the heat released by the system and the work done by the system, (d) the heat released by the system and the work done on the system.

Determine the Concept According to the first law of thermodynamics, the change in the internal energy of the system is equal to the heat that enters the system plus the work done on the system. (b) is correct.

11 ? [SSM] A real gas cools during a free expansion, while an ideal gas does not cool during a free expansion. Explain the reason for this difference.

Determine the Concept Particles that attract each other have more potential energy the farther apart they are. In a real gas the molecules exert weak attractive forces on each other. These forces increase the internal potential energy during an expansion. An increase in potential energy means a decrease in kinetic energy, and a decrease in kinetic energy means a decrease in translational kinetic energy. Thus, there is a decrease in temperature.

12 ? An ideal gas that has a pressure of 1.0 atm and a temperature of 300 K is confined to half of an insulated container by a thin partition. The other half of the container is a vacuum. The partition is punctured and equilibrium is quickly reestablished. Which of the following is correct? (a) The gas pressure is 0.50 atm and the temperature of the gas is 150 K. (b) The gas pressure is 1.0 atm and the temperature of the gas is 150 K. (c) The gas pressure is 0.50 atm and the temperature of the gas is 300 K. (d) None of the above.

1690 Chapter 18

Determine the Concept Because the container is insulated, no energy is exchanged with the surroundings during the expansion of the gas. Neither is any work done on or by the gas during this process. Hence, the internal energy of the gas does not change and we can conclude that the equilibrium temperature will be the same as the initial temperature. Applying the ideal-gas law for a fixed amount of gas we see that the pressure at equilibrium must be half an atmosphere. (c)

is correct.

13 ? A gas consists of ions that repel each other. The gas undergoes a free expansion in which no heat is absorbed or released and no work is done. Does the temperature of the gas increase, decrease, or remain the same? Explain your answer.

Determine the Concept Particles that repel each other have more potential energy the closer together they are. The repulsive forces decrease the internal potential energy during an expansion. A decrease in potential energy means an increase in kinetic energy, and an increase in kinetic energy means an increase in translational kinetic energy. Thus, there is an increase in temperature.

14 ? Two gas-filled rubber balloons that have equal volumes are located at the bottom of a dark, cold lake. The temperature of the water decreases with increasing depth. One balloon rises rapidly and expands adiabatically as it rises. The other balloon rises more slowly and expands isothermally. The pressure in each balloon remains equal to the pressure in the water just next to the balloon. Which balloon has the larger volume when it reaches the surface of the lake? Explain your answer.

Determine the Concept The balloon that expands isothermally is larger when it reaches the surface. The balloon that expands adiabatically will be at a lower temperature than the one that expands isothermally. Because each balloon has the same number of gas molecules and are at the same pressure, the one with the higher temperature will be bigger. An analytical argument that leads to the same conclusion is shown below.

Letting the subscript a denote the adiabatic process and the subscript i denote the isothermal process, express the equation of state for the adiabatic balloon:

P0V0

=

PfVf,a Vf,a

=

V0

P0 Pf

1

For the isothermal balloon:

P0V0

= PfVf,i

Vf,i

=

V0

P0 Pf

Heat and the First Law of Thermodynamics 1691

Divide the second of these equations by the first and simplify to obtain:

Vf,i Vf,a

=

V0 V0

P0 Pf

P0 Pf

1

=

P0 Pf

1-1

Because P0/Pf > 1 and > 1:

Vf,i > Vf,a

15 ? A gas changes its state quasi-statically from A to C along the paths shown in Figure 18-21. The work done by the gas is (a) greatest for path ABC, (b) least for path AC, (c) greatest for path ADC, (d) The same for all three paths.

Determine the Concept The work done along each of these paths equals the area under its curve. The area is greatest for the path ABC and least for the path ADC. (a) is correct.

16 ? When an ideal gas undergoes an adiabatic process, (a) no work is done by the system, (b) no heat is transferred to the system, (c) the internal energy of the system remains constant, (d) the amount of heat transferred into the system equals the amount of work done by the system.

Determine the Concept An adiabatic process is, by definition, one for which no heat enters or leaves the system. (b) is correct.

17 ? True or false:

(a) When a system can go from state 1 to state 2 by several different processes, the amount of heat absorbed by the system will be the same for all processes.

(b) When a system can go from state 1 to state 2 by several different processes, the work done on the system will be the same for all processes.

(c) When a system can go from state 1 to state 2 by several different processes, the change in the internal energy of the system will be the same for all processes.

(d) The internal energy of a given amount of an ideal gas depends only on its absolute temperature.

(e) A quasi-static process is one in which the system is never far from being in equilibrium.

(f) For any substance that expands when heated, its Cp is greater than its Cv.

1692 Chapter 18

(a) False. The amount the internal energy of the system changes along any path depends only on the temperatures at state 1 and state 2, but because Eint = Qin +Won , Qin depends on Won, which, in turn, depends on the path taken from state 1 to state 2.

(b) False. The work done on the system depends on the path taken from state 1 to state 2.

(c) True. The amount the internal energy of the system changes along any path depends only on the temperatures at state 1 and state 2.

(d)

True.

For

an

ideal

gas, Eint

=

3 2

nRT

.

For

a

given

amount

of

an

ideal

gas,

n

is

constant.

(e) True. This is the definition of a quasi-static process.

(f) True. All materials have values for Cp and Cv for which it is true that Cp is greater than Cv. For liquids and solids we generally ignore the very small difference between Cp and Cv .

18 ? The volume of a sample of gas remains constant while its pressure increases. (a) The internal energy of the system is unchanged. (b) The system does work. (c) The system absorbs no heat. (d) The change in internal energy must equal the heat absorbed by the system. (e) None of the above.

Determine the Concept For a constant-volume process, no work is done on or by the gas. Applying the first law of thermodynamics, we obtain Qin = Eint. Because the temperature must change during such a process, we can conclude that

Eint 0 and hence Qin 0. (d ) is correct.

19 ? When an ideal gas undergoes an isothermal process, (a) no work is done by the system, (b) no heat is absorbed by the system, (c) the heat absorbed by the system equals the change in the system's internal energy, (d) the heat absorbed by the system equals the work done by the system.

Determine the Concept Because the temperature does not change during an isothermal process, the change in the internal energy of the gas is zero. Applying the first law of thermodynamics, we obtain Qin = -Won = Wby the system. Hence

(d ) is correct.

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