CHANGE OF STATE - Antimatter



CHANGE OF STATE

On heating

SOLID LIQUID GAS

The change of state from a solid to a liquid takes place at a constant temperature known as the MELTING POINT.

The change of state from a liquid to a gas or vapour takes place at a constant temperature known as the BOILING POINT.

WHEN CHANGING STATE THERE IS NO CHANGE IN TEMPERATURE.

e.g

ICE WATER AT 0OC

WATER STEAM AT 100OC

Therefore heat energy CANNOT produce BOTH a change of state and a change in temperature simultaneously. At any given time the material is EITHER changing temperature or changing state.

The specific latent heat of fusion is defined as the heat energy required to change unit mass of a substance from the solid to the liquid state at its melting point.

Symbol l f

The specific latent heat of vapourisation is defined as the quantity of heat required to change unit mass of a substance from the liquid to the vapour state without a change in temperature i.e at the boiling point

Symbol lv

[pic]Rearranging the equation gives [pic]

The equation for heat energy required to change the state of m kg of the substance at a constant temperature.

[pic] Solid to Liquid at m.p

[pic] Liquid to gas at b.p.

lf ( water ) = 3.35 x 105 J kg-1

lv ( water ) = 2.26 x 106 J kg-1

How much energy is required to raise the temperature of 1 kg of water by one degree ?

How much energy is required to convert 1 kg of ice at 0o C to 1 kg of water at 0oC ?

How much energy is required to convert 1 kg of water at 100o C to 1 kg of steam at 100oC ?

Example : Calculate the amount of heat energy required to change 20 g of ice at

-20oC into steam at 100oC

Given that

s.h.c. ( water ) = 4,200 J kg-1 K-1

s.h.c. ( ice ) = 2,100 J kg-1 K-1

lf = 3.35 x 105 J kg-1

lv = 2.26 x 106 J kg-1

Remembering that the melting point if ice is 0o and the boiling point of water is 100o we need to divide this question into four stages.

Stage 1. Calculate the energy required to change 20 g of ice at –20oC to ice at 0o

This is a change of temperature

Q = m c Δ T

Q = 20 x10-3 x 2,100 x 20 = 840 J

Stage 2. Energy required to change 20 g of ice at 0oC to water at 0oC

This requires a change of state solid to liquid

Q = m lf

Q = 20 x 10-3 x 3.35 x 105 = 6700 J

Stage 3. Energy required to change 20 g of water at 0oC to water at 1000C.

This is a change of temperature

Q = m c ΔΤ

Q = 20x10-3 x 4,200 x 100 = 8400 J

Stage 4. Energy required to change 20 g of water at 100oC to steam at 100oC.

This is a change of state liquid to gas

Q = m lv

Q = 20 x 10-3 x 2.26 x 106 = 45200 J

Therefore the total energy required is equal to

840 + 6700 + 8400 + 45200 = 6.114 x 104 J

Experiment to Measure the Specific latent heat of fusion of water

Q = m l f

where lf = specific latent heat of fusion, m = mass of the substance and

Q = the quantity of heat energy supplied.

In the experiment the heat energy is supplied by an electrical heater and therefore the energy supplied

Q = V I t

where V = voltage, I = current, and t = time of heating.

It is difficult to prevent heat losses from the system, but we can account for them as follows

V I t = m lf + losses ........................ 1

If the experiment is repeated with different values for voltage V1 and current I 1 but for the same time t then the mass that changes state is m 1

V1 I 1 t = m1l f + losses ................... 2

However the losses should be the same in each case so subtracting equation 1 from equation 2 gives

V I t - V1 I 1 t = m l f - m1 lf

Rearranging

[pic]

Experimental Method

1. Set up a heating circuit and a control circuit as shown, by placing ice into two funnels held over two beakers. Place a heater into one of the funnels.

V V

A

CONTROL HEATING EXPERIMENT

2. Heat the ice for 600 seconds ensuring that the heater is always covered fully by ice.

3. Note the value for the current and voltage for the heating circuit.

4. Calculate the mass of ice melted by the electrical heater alone by subtracting the mass of ice melted in the control experiment in the same time.

5. Repeat the experiment, using a different heater voltage and current but the same heating time.

6. From the results calculate the specific latent heat of fusion for water.

Problem Sheet 3 SPECIFIC LATENT HEAT

REMEMBER ALWAYS USE THE EQUATION

HEAT LOST = HEAT GAINED

WHEN MIXING SUBSTANCES AT DIFFERNET TEMPERATURES

Question 1 A heat transfer of 9.0×105 J is required to convert a block of ice at -20oC

to water at 15oC .What was the mass of the block of ice?

Question 2. A 300 g copper object is dropped into a 150 g copper calorimeter containing 220 g of water at 20oC. It causes the water to boil and 5 g is converted into steam. Calculate the original temperature of the copper object.

Question 3. A 1.1 kg block of ice is initially at a temperature of -2.0oC .If 2.4×105 J of heat are added to the ice, what is the final temperature of the system? Find the amount of ice, if any, that remains.

Question 4. A 2 kg insulated bar of steel of length 150 mm is supplied with heat energy at a rate of 100 W. Calculate the increase in length of the bar after 10 minutes.

s.h.c. (steel) = 500 J kg-1 K –1 α coefficient of linear expansion(steel) = 1.2 x 10-5 oC –1

Question 5.(a) How much energy is required to convert 1500 g of ice at –35oC into steam at 100oC (b)If the heat is being supplied at a rate of 4.5 kW how long will it take?

Question 6. A thin walled copper cylindrical flask of mass 170 g which provides perfect insulation and is fitted with a 3 kW heater and filled with water.

The dimensions of the flask are 8 cm diameter and height 12 cm

a) How long does it take for the heater of raise the temperature of the water from 18oC to 100oC

b) If the heater is left on how far will the water level drop in 4 minutes

c) How long will it take for the flask to boil dry

Question 7. (a) Calculate the heat released as 20g of liquid water at 100oC is cooled to 60 oC.(b) Calculate the heat released as 20 g of liquid water at 100 oC is cooled to 60 oC (c) Find the mass of flesh that can be heated from 37oC (normal body temperature) to 60 oC for the case considered in part A. (The average specific heat of flesh is 3500 Jkg-1K-1.)

d) Find the mass of flesh that can be heated from 37oC (normal body temperature) to 60 oC for the case considered in part B. (The average specific heat of flesh is 3500 Jkg-1K-1.)

HEATING CURVES / COOLING CURVES

A heating curve is a plot of temperature ( Y AXIS ) versus time ( X AXIS ) taken when heating the substance.

A cooling curve is a plot of temperature ( Y AXIS ) versus time( X AXIS ) when the substance is being cooled

Heating / Cooling curves are used to determine the melting point or boiling point of a substance.

The melting or boiling points are the points at which the graph levels off

Example : Heating curve for water

Plot a graph of temperature versus time for water as it is being heated.

100ob.p. Vapour steam

Temp

( o C )

liquid water

0o m.p solid ice

Note

1. Where the graph levels off gives the values for the melting points and boiling points for water.

2. The length of the level line (i.e the time ) representing the conversion from liquid to gas is approximately ten time the length of the line

( again the time value ) for the conversion from solid to liquid this is the case since

lv = 2.26 x 106 J / kg and lf = 3.35x105 J / kg

Approximately a factor of ten

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