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, subtract the initial temperature of the water + calorimeter 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 found by using the formula for the mass of calorimeter and water combined, where the total mass is the known mass of the calorimeter and the cold water. Specifically, the mass can be determined by weighing both the calorimeter and cold water together and then using the mass of the calorimeter separately to subtract and find the mass of the cold water.

The steam was dried by using a steam trap or delivery tube sloping upwards to allow any condensed steam or water to drain away, thus ensuring that only dry steam enters the calorimeter.

The heat gained by the water and the calorimeter can be calculated using the formula:

$Q = mc heta$

Where:

• $m$ is the mass of the water (67.50 g) plus the mass of the calorimeter (73.40 g), converted into kg.
• $c$ is the specific heat capacity of water (4180 J/kg·K).
• $heta$ is the rise in temperature (9 °C).

Total mass = (67.50 g + 73.40 g) ÷ 1000 = 0.1409 kg.

Calculating the heat:

$Q = 0.1409 imes 4180 imes 9 = 5294.0 J$.

To find the heat lost by the steam, the mass of steam (1.03 g) should be converted to kg and used in conjunction with the specific heat capacity:

$Q = mc heta$

Where:

• $m = 1.03 g ÷ 1000 = 0.00103 kg$ (mass of the steam)
• $c = 4180 J/kg·K$ (specific heat capacity of water)
• $heta = 100 °C - 19 °C = 81 °C$ (change in temperature)

Calculating the heat:

$Q = 0.00103 imes 4180 imes 81 = 348.7 J$.

Using the heat values calculated earlier, we can express the latent heat of vaporization using:

$L = \frac{Q_{lost} - Q_{gained}}{m_{steam}}$

Substituting the values:

• $Q_{gained} = 5294.0 J$
• $Q_{lost} = 348.7 J$
• $m_{steam} = 1.03 g = 0.00103 kg$

Thus,

$L = \frac{(5294.0) - (348.7)}{0.00103} = 2448 J/kg$,

And in significant figures, this can be expressed as: $L = 2.37 imes 10^3 J/kg.$