2 Latent heat

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Specific Latent Heat

• In Thermodynamics ! We saw that energy is needed to break inter-atomic attractions when a substance melts or boils.

• This energy is called latent heat.• The temperature is constant during this change of

state.• The following equation is used to calculate energy

needed for a change of state.

Heat transferred = mass x specific latent heat capacity

ΔQ (J) = m (kg) x L (J kg-1)

Specific latent heat, L is the energy needed to change the state of 1 kg of the substance (without change in temperature).

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Specific latent heat• Latent heat of fusion refers to the change from a

solid to a liquid (melting)• Latent heat of vaporisation refers to the change

from a liquid to a gas (boiling).• Example 1 – the specific latent heat of fusion of

ice is 330 000 J kg-1. How much energy is needed to melt 0.65 kg of ice?

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ΔQ = mL = 0.65 kg x 330 000 J kg-1

= 210 000 J

Measuring specific latent heat• Look at the diagram in example 2.• The water is boiling and stays at a constant 100 oC• Any energy delivered from the heater will turn the

water into steam.• Knowing the power of the heater, • the energy delivered (J) = power (W) x time (s)• Example 2: the power of the emersion heater is

60 W. In 5 minutes, the top pan balance reading falls from 282 g to 274 g. calculate the specific latent heat of vaporisation of water.

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Measuring specific latent heat• Energy, ΔQ = power x time• = 60 W x (5 x 60) s = 18 000 J• Mass of water evaporated = 282 g – 274 g• = 8 g = 8 x 10-3 kg• ΔQ = mL hence L = ΔQ / m• = 18 000 J / 8 x 10-3 kg = 230000 J kg-1. • A similar method can be used for finding the

latent heat of fusion.

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Funnel is filled with ice at 0 oC, and the mass of water produced in 5 minutes is noted.

In a warm room, some ice will melt with energy received from the room.

Measuring specific latent heat• To allow for this, experiment is repeated twice

for same length of time.• Once with heater off and once with it on.• First experiment is the control – indicates how

much ice melts by absorbing energy from the room.

• Ice melted by heater is the difference between the masses in the two experiments.

Specific heat capacity• For a substance being heated, the rise in

temperature depends on:1. Mass of substance being heated2. How much energy has been put in3. What the substance is

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Specific heat capacity• More energy is needed to the same temperature

rise to 1 kg of water than 1 kg of copper.

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ΔU = m x c x ΔTΔU is the change in internal energy (J)m is the mass of substance (kg)c is the specific heat capacity (J kg-1 K-1)ΔT is the temperature change (K)

If the energy is supplied as heat, equation can be written as: ΔQ = m x c x ΔT

Specific heat capacity depends on the substanceWater has a value of 4200 J kg-1 K-1, while copper hasA value of 380 J kg-1 K-1.

Specific heat capacity• Defined as the energy needed to raise the

temperature of 1 kg of substance by 1 K.• Example 3: 0.5 kg of water is heated from 10 oC to

100 oC. How much does its internal energy rise?• Rise in temperature = 100 oC – 10 oC = 90 oC• This is also a rise of 90 K (oC and K are the same

size)• ΔU = m x c x ΔT• = 0.5 kg x 4200 J kg-1 K-1 x 90 K• = 190 000 J.Measuring specific heat capacity• An experiment can be done to find the specific heat

capacity using an emersion heater.

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Worked example 4In an experiment shown in the diagram below, a 60 W immersion heater was used. The beaker contained 1 kg of water at 21 oC. After 5 minutes, the heater was switched off. The temperature of the water went up to 25 oC. What is the specific heat capacity of the water?

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to power supply thermometer

1 kg of water

immersion heater

Solution

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Change in internal energy, ΔU = power x time = 60 W x (5 x 60) s

= 18 000 J

Rise in temperature = 25 oC – 21 oC = 4 oC = 4 K

ΔU = m.c.ΔT 18 000 J = 1 kg x c x 4 K

hence c =

18 000 J 4 K x 1 kg

= 4500 JKg-1K-1

The result is too large. Can you see why?

The 18 000 J from the electrical heater does not all go to increase the internal energy of the water.Some heat will escape to increase internal energy of the room, there is an increase in the rest of the apparatus.

Energy losses in measuring specific heat capacities

The increase in internal energy of water is less than 18 000 J, and can be calculated from: ΔU = m.c.ΔT = 1.0 kg x 4200 J kg-1 x 4 K

= 17 000 JThe extra 1000 J supplied by the heater heats up the apparatus and the surroundings.

Comparing specific heat capacitiesSpecific heat capacities can also be determined for metal blocks of different materials.To reduce heat losses, metal blocks are lagged with insulating materialTable on slide 11 gives the specific heat capacities of 2 metals, measured this way

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Substance Aluminium Iron

c / J kg-1 K-1 880 470

1 kg of Al needs almost twice as much energy as 1 kg of Fe for each 1 K rise – why so?

If number of atoms of Al and Fe are the same, energy needed will be similar

Al atom is about half the mass of Fe atom, so contains 2 x as many atoms per kg as Fe

For comparison, the number of atoms used is the Avogadro constant (6 x 1023 atoms), which makes 1 moleThe molar heat capacity is the energy needed to raise the temperature of one mole of a metal by 1 K.

Molar heat capacities are similar – same number of atoms

Questions 1. A horseshoe of mass 0.8 kg is heated from 20 oC to 900

oC. How much does its internal energy rise? (The specific heat capacity of steel is 470 J kg-1 K-1).

2. A kilogram of water falls 807 m (the height of the Angel Falls in Venezuela) and gains 7900 J of kinetic energy. Assuming that this is all converted to internal energy in the water, calculate the rise in temperature of the water. The specific heat capacity of the water is 4200 J kg-1 K-1.

3. A kettle contains 1.5 kg of water at 18 oC.(a) How much heat energy is needed to raise the temperature

of the water to 100 oC?(b) Assuming no energy is lost to the room, calculate how

long this will take if the power rating of the kettle is 2000 W.

(c) What else must you assume in part (b)?The specific heat capacity of water is 4200 J kg-1 K-1.

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4. Once the kettle in question 3 has reached 100 oC, the water will boil. Assume the lid has been left off, so the kettle does not switch itself off.

(a) How much energy is needed to boil away 0.5 kg of water? (the specific latent heat of vaporisation of water, L = 2.3 x 106 J kg-1).

(b) Assuming no energy is lost, calculate how this will take if the power rating of the kettle is 2000 W).

5. The specific latent heat of fusion of ice is 330 000 J kg-1.(a) Calculate the energy needed to melt 50 g of ice at its

melting point.(b) A galss contains 0.4 kg of lemonade at 20 oC, and 50 g of

ice at 0 oC are put in to cool it. Show that the temperature of the lemonade drops to about 10 oC.(You assume that the specific heat capacity of lemonade is 4200 J kg-1 K-1 and that the energy needed to melt the ice all comes from the lemonade).