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
Specific latent heat 
l 
Heat required to change the state of the substance
mass of the substance
Q
m
S.I. Units
Joules
 J .kg 1
kg
Rearranging the equation gives
Q  m.l
The equation for heat energy required to change the
state of m kg of the substance at a constant
temperature.
Q  m.l f
Solid to Liquid at m.p
Q  m.lv
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 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 
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 lf
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 = VIt
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
lf 
VI  V1 I1 t
m  m1
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.
Power
Supply
VV
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  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
(oC)
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