Latent Heat


Physics Lab Manual

Part 1: Heat of fusion

When water molecules are packed into a solid lattice they form bonds with their nearest neighbor molecules. To “melt” the solid (ice) requires the addition of energy. If the ice is at 0°C the addition of heat serves to break these bonds between the water molecules into a liquid state. Thus, when heat is added to the ice it melts rather than increasing in temperature. In this sense, the addition of heat is not directly noticeable (by a rise in temperature) and is termed “latent” heat. Such a situation is characteristic also of liquid being turned into vapor. The latent heat of fusion (associated with the melting of a solid) is defined by:

Lf = Heat required to melt 1g of solid

Thus, to melt a mass m of ice it takes an amount of energy Q = mLf. The task of this laboratory will be to determine Lf for water using calorimetry. The idea is to use our knowledge of the specific heat of water and a thermometer to determine how much heat is absorbed when a given amount of ice is melted in water. Knowing the mass of ice it’s then straightforward to find Lf.


  1. Using the scale at the front of the room determine the mass of the calorimeter.

  2. Fill the calorimeter roughly halfway full with room temperature water. Determine both the mass of the water and its temperature.

  3. Add enough ice to lower the mixture to 5° C. Make sure all the ice is melted and to stir the mixture when assessing its final temperature.

  4. Determine the mass of ice that was added to the mixture.


The key to understanding how this problem works out is to regard the water initially in the calorimeter and the ice as two separate systems. Imagine that the ice was placed in a plastic bag and then dropped in the water (of course this would make no difference).

  1. Using your measurements for the initial mass and temperature of the water determine how much heat is lost by the water initially in the calorimeter (recall Q = mcDT where c = 1cal/g°C). You should be able to get a number for this heat.

  2. Write down an expression for how much heat is absorbed by the ice. Note that the ice must first be melted and then heated up. Your expression should involve the unknown Lf.

  3. Use the principle of conservation of energy to relate the two expressions you found in parts (1) and (2) and solve for the latent heat of ice. Compare with the accepted value of 79.7cal/g.


  1. Is heat lost or absorbed from the environment by the ice-water mixture? How will this affect your result for the latent heat? Will it be systematically high or low?

  2. If you had the choice to use either cubes of ice or shaved ice what might yield a more accurate value for the latent heat? Explain.

Part 2: Heat of vaporization

Like the case of melting, to vaporize a liquid requires the breaking of bonds between molecules. There is correspondingly a latent heat of vaporization.


  1. Put 400g of water into a metal canister and measure its original temperature.

  2. Get a hot plate and preheat it for 5 minutes.

  3. Place the canister full of water on the hotplate and after 2 minutes record its temperature. Continue to keep track of how long the water is on the hot plate.

  4. Bring the water to a boil and boil it for an additional 5 minutes and then remove the canister from the hot plate. Record the total time the canister was on the hot plate and the mass of water that has evaporated.


  1. Determine the change in temperature of the water after being on the hot plate for 2 minutes and from this the amount of heat transferred to the water from the hotplate during these two minutes. At what rate (cal/min) does the hotplate transfer heat to the water?

  2. Use the rate that heat is transferred to the water and the total time the water was on the hot plate to determine the total heat transferred to the water by the hotplate. Call this Qhp.

  3. The heat absorbed by the water goes into raising the temperature of the water and converting some of it into steam. Write down an expression for this absorbed heat. Call it Qw. Your expression for Qw should involve the unknown heat of vaporization Lv­..

  4. Use energy conservation to relate Qhp to Qw and from this determine an experimental value for Lv. Compare with the accepted value of Lv = 540cal/g.


Is your value for Lv too small or too large? Can you offer some explanation for the discrepancy in terms of the energy lost from the water to the environment?