In an experiment to measure the specific latent heat of vaporisation of water, cool water was placed in an insulated copper calorimeter - Leaving Cert Physics - Question 2 - 2005

To calculate the specific latent heat of vaporisation of water ( L_v), we first determine the heat gained by the water in the calorimeter and the calorimeter itself.

Step 1: Calculate the change in temperature

The initial temperature of the water is 10 °C and the final temperature is 25 °C:

$\Delta T = T_f - T_i = 25 °C - 10 °C = 15 °C$

Step 2: Calculate mass of water

Mass of the water can be calculated using:

$\text{Mass of water} = \text{Mass of calorimeter + water} - \text{Mass of calorimeter} = 91.2 g - 50.5 g = 40.7 g = 0.0407 kg$

Step 3: Calculate heat gained by the system

Using the formula for heat gain:

$Q = mc\Delta T$

Where:

• m = mass of water (0.0407 kg)
• c = specific heat capacity of water (4200 J kg⁻¹ K⁻¹)
• \Delta T = 15 °C

Thus:

$Q = 0.0407 \, \text{kg} \times 4200 \, \text{J} \text{kg}^{-1} \text{K}^{-1} \times 15 \, \text{K}$

Calculating:

$Q = 0.0407 \times 4200 \times 15 = 2564.1 \, \text{J}$

Step 4: Calculate heat lost by steam

The lost heat by steam is equal to the heat gained by the water and the calorimeter combined.

For the calorimeter:

• Mass of calorimeter = 50.5 g = 0.0505 kg
• Specific heat capacity of copper = 390 J kg⁻¹ K⁻¹

Heat gained by the calorimeter:

$Q_{cal} = mc\Delta T$

$Q_{cal} = 0.0505 \, \text{kg} \times 390 \, \text{J} \text{kg}^{-1} \text{K}^{-1} \times 15 \, \text{K} = 295.425 \, \text{J}$

Step 5: Total heat gained

Total heat gained:

$Q_{total} = Q_{water} + Q_{cal} = 2564.1 \, \text{J} + 295.425 \, \text{J} = 2859.525 \, \text{J}$

Step 6: Calculate specific latent heat of vaporisation

Since the heat lost by the steam (L_v) is equal to the total heat gained:

$L_v = \frac{Q_{total}}{m_{steam}}$

Where:

• $m_{steam}$ = mass of steam added = $\text{Mass of calorimeter + water + steam} - \text{Mass of calorimeter + water} = 92.3 g - 91.2 g = 1.1 g = 0.0011 kg$

Thus:

$L_v = \frac{2859.525 \, \text{J}}{0.0011 \, \text{kg}} = 2608650.45 \, \text{J kg}^{-1} = 2.61 \times 10^6 \, \text{J kg}^{-1}$