I. Title Latent Heat of Fusion
Introduction: Most homogeneous solids, including ice, have a very specific melting temperature. You
know that pure ice melts at __0_oC at atmospheric pressure. When a sample of ice is melting to form
water, the entire ice/water mixture remains at _0__oC until all of the ice is melted. This is evident when
you are drinking a glass of ice water. The water really doesn't “heat up” until all of the ice is gone.
The Latent heat of fusion of a solid is the amount of energy that must be transferred to one gram of the
solid, at its melting point temperature, in order to melt it. It is called latent heat because there is no
temperature change associated with this energy transfer, there is only a change in phase.
Specific heat capacity is the quantity of heat required to raise the temperature of a unit of mass of a
substance by a unit change in temperature.
Q = mc  T
where Q is the change in heat transferred in Joules, m is the mass in g, c is the specific heat
capacity in cal/ g oC or J/(g°C), and T is the change in transferred in °C or K.
The derived unit for c, the specific heat capacity, is cal/ g oC or J/(g°C),
The specific heat capacity of a substance depends on its atomic /molecular structure and on its phase(s,l,g)
The specific latent heat of a substance is the quantity of heat energy required to change the state of a unit
mass of a substance.
QL= mL where QL is the heat transferred, in joules, m is the mass, in g, and L is the latent heat in joules
per gram.
The SI unit for specific latent heat is J/kg.
The specific latent heat of fusion is the quantity of heat energy released when 1 g of a substance solidifies
(i.e. fuses) without changing its temperature.
The specific latent heat of vaporization is the quantity of heat energy needed to “boil” 1 g of a substance
without changing its temperature.
Water has one of the largest specific latent heats of fusion of all substances.
Research Question:
In this experiment you will determine the latent heat of fusion of ice by melting a known mass of ice in
some warm water. An energy balance on the system will reveal the heat of fusion.
III. Method:
You will need: Styrofoam cup, warm water, thermometer, ice, balance
1. a) Determine the mass of a Styrofoam cup
b) Add approxiamately 30 mL water to the cup.
c) Determine the mass of the Styrofoam cup and water.
d) Determine the initial temperature of the water before the ice is added.
2. Place one or two pieces of ice that has been patted dry with a paper towel into the stryofoam cup
and stir with the thermometer until the ice melts.
3. Record the lowest temperature the water in the Styrofoam cup achieves.
4. Determine the new mass of the water in the Styrofoam cup.
IV. Results Table: The following data can be used to determine the latent heat of fusion of ice.
mass of Styrofoam cup = 10.26g
mass of cup and room temperature water 77.24g
initial temperature of the room temperature water
23.0oC
equilibrium temperature of the water after the ice melts 4.5oC
mass of calorimeter and water after the ice has melted. 85.86 g
V. Conclusions:
Q = m c (Tfinal-Tinitial)
c= 4.180 J / goC of water c= 1 cal / goC
1. Determine the heat lost by the room temperature water in calories.
Heat = mass of water ( temp change ) ( specific heat capacity )
Q = m (Tfinal-Tinitial) c = ( 77.24g – 10.26g)( 4.5oC-23.0oC) (1 cal / goC)= 791 cal = 7.9x102 cal
2. After the ice melts it is liquid water with a temperature of 0 oC and a specific heat of 1cal/goC. or
4.180 J / goC.
This water gains energy until it reaches the equilibrium temperature.
Determine the amount of heat gained by the ice ( water after it melts) in calories.
Heat = mass of ice water ( temp change ice water ) ( specific heat capacity of water)
Q = m c (Tfinal-Tinitial) = ( 85.86g – 77.24g)( 4.5oC-0.0oC) (1 cal / goC)= 39 cal
3. Using the principle of conservation of energy determine the amount of heat absorbed by the
melting ice in cal and joules
7.9x102 cal – 39 cal = 751 cal = 7.5x102 cal
Heat lost by water-Heat gained by ice water = Latent or hidden heat needed to melt the ice.
4. Determine the Latent Heat of Fusion of Ice in cal per g and J per kg
Hf = 7.5 x102 cal = 87 cal/g
87cal( 4.18 J) ( 1000g ) = 3.6 x106 J / kg
( 85.86g – 77.24g)
g ( 1cal ) ( 1 kg )
mass of ice melted