In an experiment to measure the specific latent heat of vaporisation of water, a student used a copper calorimeter containing water and a sensitive thermometer - Leaving Cert Physics - Question 2 - 2010

The steam was dried using a steam trap or delivery tube that sloped upwards. This design allows for the condensation process to occur by gravity, ensuring that only dry steam enters the calorimeter, preventing the introduction of excess water that could alter the results.

The mass of the steam was determined by weighing the calorimeter and its contents before and after adding the steam. The initial mass of the calorimeter and water is subtracted from the final mass, which includes the calorimeter, water, and the condensed steam. Therefore, the mass of the steam is given by:

$m_s = (m_{final} - m_{initial}) - m_{well}$ where $m_s$ is the mass of the steam, $m_{final}$ is the mass after adding steam, and $m_{initial}$ is the mass of the calorimeter and water before adding the steam.

A sensitive thermometer was used to provide accurate and precise temperature readings, allowing for better detection of small temperature changes. This is particularly important in experiments measuring thermal properties because even minor fluctuations in temperature can greatly influence the calculations of specific latent heat.

To calculate the specific latent heat of vaporisation of water, we start with the equation of heat gained by the calorimeter and water:

$Q_{water} = m_s L_v$

Where:

• $Q_{water}$ is the heat gained by the water,
• $m_s$ is the mass of the steam,
• $L_v$ is the latent heat of vaporisation.

Given:

• Mass of dry steam, $m_s = 1.2 imes 10^{-3} ext{ kg}$
• Initial temperature of cooled water = 8.2 °C, Final temperature = 20.0 °C
• Specific heat capacity of water, $c_w = 4180 ext{ J kg}^{-1} ext{ K}^{-1}$
• Specific heat capacity of copper, $c_c = 390 ext{ J kg}^{-1} ext{ K}^{-1}$
• Mass of copper calorimeter, $m_c = 34.6 ext{ g} = 0.0346 ext{ kg}$

Calculating the heat absorbed by water:

$Q_{water} = (m_{water}) c_w (T_{final} - T_{initial})$
Substituting;

$m_{water} = 96.4 ext{ g} - 34.6 ext{ g} = 61.8 ext{ g} = 61.8 imes 10^{-3} ext{ kg}$

Therefore:

$Q_{water} = (0.0618) (4180)(20.0 - 8.2) = 11.8 ext{ kJ}$

Now calculating the specific latent heat using:

L_v = rac{Q_{water}}{m_s} = rac{11800 ext{ J}}{1.2 imes 10^{-3} ext{ kg}} = 9840000 ext{ J kg}^{-1} = 2.34 imes 10^6 ext{ J kg}^{-1}