During an experiment to measure the specific latent heat of vaporisation of water, cold water was placed in an insulated copper calorimeter - Leaving Cert Physics - Question 2 - 2011

To find the rise in temperature of the water, subtract the initial temperature of the water from the final temperature:

Rise in temperature = Final temperature - Initial temperature
Rise in temperature = 19 °C - 10 °C = 9 °C

The mass of the cold water was calculated by weighing the calorimeter filled with cold water. The mass of the empty calorimeter is known, and then the total mass (calorimeter + water) is measured. The mass of the water can be found by subtracting the mass of the empty calorimeter from the total mass recorded.

The steam was dried using a steam trap or drying tube that is positioned to ensure that the steam enters only as dry steam into the calorimeter. The delivery tube is sloped upwards to prevent condensation of steam inside the tube before reaching the calorimeter.

The heat gained can be calculated using the formula:

$Q = (m_{water} + m_{cal}) imes c_{water} imes riangle T$
Where:

• $m_{water} = 67.50 ext{g}$
• $m_{cal} = 73.40 ext{g}$
• $c_{water} = 4180 ext{ J kg}^{-1} ext{K}^{-1}$
• $riangle T = 9°C$

Converting the mass to kg:

• $m_{water} = 0.0675 ext{ kg}$
• $m_{cal} = 0.0734 ext{ kg}$

Calculating: $Q = (0.0675 + 0.0734) imes 4180 imes 9 = 2539.4 ext{ J}$

The heat lost by the condensed steam can be found using the formula:

$Q = m_{steam} imes L_{v}$
Where:

• $m_{steam} = 1.03 ext{ g}= 0.00103 ext{ kg}$
• $L_{v} = 4180 ext{ J kg}^{-1}$

Calculating: $Q = 0.00103 imes 4180 = 4.296 ext{ J}$

Using the heat gained by water and calorimeter and the heat lost by the steam:

First, recall: $Q_{gained} = Q_{lost}$
Thus, L_{v} = rac{Q_{gained}}{m_{steam}}

Substituting: L_{v} = rac{2539.4}{0.00103} = 2467.8 ext{ J kg}^{-1}