Learn It: Experiment 2

Measurement of the specific latent heat of fusion of ice

Apparatus

Ice, water, calorimeter, lagging, beakers, kitchen paper, digital thermometer reading to 0.1°C and electronic balance.

Procedure

1. Place some ice cubes in a beaker of water and keep taking the temperature with the thermometer until the ice-water mixture reaches 0 °C.
2. Find the mass of the calorimeter m_cal.
3. Half fill the calorimeter with water warmed to approximately 10 °C above room temperature.  Find the combined mass of the calorimeter and water m₂.  The mass of the water m_w is m₂ – m_cal.
4. Record the initial temperature q₁ of the calorimeter plus water.
5. Surround the ice cubes with kitchen paper or a cloth and crush them between wooden blocks – dry them with the kitchen paper.
6. Add the pieces of dry crushed ice, a little at a time, to the calorimeter.  Do this until the temperature of the water has fallen by about 20 °C.
7. Record the lowest temperature q₂ of the calorimeter plus water plus melted ice.  The rise in temperature of the ice Dq₁ is q₂  – 0°C, and the fall in temperature of the calorimeter plus water Dq₂  is q₁ – q₂.
8. Find the mass of the calorimeter plus water plus melted ice m₃.  The mass of the melted ice m_i is m₃ – m₂.

Results
Mass of the calorimeter m_cal =
Mass of the calorimeter plus water m₂ =
Mass of the water m_w = m₂ – m_cal =
Initial temperature of the calorimeter plus water q₁ =
Final temperature of the calorimeter plus water plus melted ice q₂ =
Rise in temperature of the ice = q₂ – 0°C =D∆q₁
Fall in temperature of the calorimeter plus water = q₁ – q₂ =D∆q₂
Mass of the calorimeter plus water plus melted ice m₃ =
Mass of the melted ice m_i = m₃ – m₂ =

Sample results
Mass of empty calorimeter = 0.0772 kg
Mass of calorimeter and water = 0.1702 kg
Mass of water = 0.0930 kg
Mass of calorimeter + water + melted ice =  0.1894 kg
Mass of melted ice (same as mass of ice) = 0.0192 kg
Specific heat capacity of water = 4180 J kg^–1K^–1
Specific heat capacity of calorimeter material = 390J kg^–1K^–1
Original temperature of ice = 0.0 °C
Initial temperature of warm water = 29.4 °C
Final temperature of mixture = 12.2 °C
Temperature rise of melted ice (ice water) = 12.2 °C
Temperature drop of calorimeter and warm water   = 17.2 °C

The ice, in melting and warming up to the final temperature, absorbs energy twice.

First:
It absorbs some to change its state from ice at 0 °C to water at 0 °C (latent heat).

Second:
It absorbs more to warm up from this newly formed water to the final temperature of the water in the calorimeter.

The energy for these comes from both the warm water and the calorimeter.

Calculations
Assume heat losses cancel heat gains.  Given that the specific heat capacity of water cw and the specific heat capacity of copper cc are already known, the latent heat of fusion of ice may be calculated from the following equation:

Energy gained by ice + energy lost by the water   = energy lost by calorimeter in melting and warming

m_il + m_ic_wDq₁ = m_calc_c + m_wc_wDq₂
0.0192l + 0.0192 × 4180 × 12.2 = 0.0930 × 4180 × 17.2 + 0.0772 × 390 × 17.2
0.0192l + 979.123 = 6686.33 + 517.86
0.0192l = 6225.06
Specific latent heat l = 324 221.88 J kg^-1
= 324 kJ kg–1
 
Next: Learn It - Experiment 3