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 by utilizing a steam trap or a delivery tube that sloped upwards, preventing the condensation of steam and allowing only dry steam to enter the calorimeter.

The mass of the steam was determined by measuring the initial mass of the calorimeter with its contents and then subtracting this from the final mass after the steam was added. Specifically, the calculation used is:

Initial mass of calorimeter + contents/water - Final mass of calorimeter and contents/water.

A sensitive thermometer was used to achieve greater accuracy in temperature measurements. It allows for a precise reduction in the temperature and is capable of reading to a very small increment, such as 0.1 °C.

Using the provided data, we can begin the calculation of the specific latent heat of vaporisation of water.

1. Calculate the mass of water:

   $m_w = 1.2 imes 10^{-3} ext{ kg}$

2. Determine the temperature change:

   $\Delta \theta = 20.0 - 8.2 = 11.8 ext{ °C}$

3. Calculate heat gained by water and calorimeter:

   $Q = m_c c_c \Delta\theta + m_w c_w \Delta\theta$

Using:

• Mass of calorimeter: $m_c = 34.6 ext{ g} = 34.6 imes 10^{-3} ext{ kg}$
• Specific heat capacity of copper: $c_c = 390 ext{ J kg}^{-1} ext{ K}^{-1}$
• Specific heat capacity of water: $c_w = 4180 ext{ J kg}^{-1} ext{ K}^{-1}$

4. Plug values into the equation to find the specific latent heat:

   $l_v = \frac{(34.6 imes 10^{-3} imes 390 imes 11.8) + (1.2 imes 10^{-3} imes 4180 imes 11.8)}{m_w}$

Simplifying the calculations will yield:

$l_v = 2.34 imes 10^6 ext{ J kg}^{-1}$