2. 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). 2
3. 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?
3
ΔQ = mL
= 0.65 kg x 330 000 J kg-1
= 210 000 J
4. Measuring specific latent heat
• Look at the diagram in example 2.
• The water is boiling and stays at a constant 100 o
C
• 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.
4
5. 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.
5
Funnel is filled with ice at 0 o
C, 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.
6. 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 heated
2. How much energy has been put in
3. What the substance is
6
7. Specific heat capacity
• More energy is needed to the same temperature rise
to 1 kg of water than 1 kg of copper.
7
ΔU = m x c x ΔT
ΔU 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 substance
Water has a value of 4200 J kg-1
K-1
, while copper has
A value of 380 J kg-1
K-1
.
8. 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 o
C to
100 o
C. How much does its internal energy rise?
• Rise in temperature = 100 o
C – 10 o
C = 90 o
C
• This is also a rise of 90 K (o
C 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.
8
9. Worked example 4
In an experiment shown in the diagram below, a 60 W
immersion heater was used. The beaker contained 1
kg of water at 21 o
C. After 5 minutes, the heater was
switched off. The temperature of the water went up
to 25 o
C. What is the specific heat capacity of the
water?
9
to power supply
thermometer
1 kg of water
immersion heater
10. Solution
10
Change in internal energy, ΔU = power x time
= 60 W x (5 x 60) s
= 18 000 J
Rise in temperature = 25 o
C – 21 o
C = 4 o
C = 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-1
K-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.
11. 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 J
The extra 1000 J supplied by the heater heats up
the apparatus and the surroundings.
Comparing specific heat capacities
Specific heat capacities can also be determined for
metal blocks of different materials.
To reduce heat losses, metal blocks are lagged with
insulating material
Table on slide 11 gives the specific heat capacities of
2 metals, measured this way
11
12. 12
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 mole
The 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
13. Questions
1. A horseshoe of mass 0.8 kg is heated from 20 o
C to 900 o
C.
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 o
C.
(a) How much heat energy is needed to raise the temperature of
the water to 100 o
C?
(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
.
13
14. 14
4. Once the kettle in question 3 has reached 100 o
C, 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 o
C, and 50 g of ice at
0 o
C are put in to cool it. Show that the temperature of the
lemonade drops to about 10 o
C.
(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).
