In an experiment to determine the specific latent heat of fusion of ice, warm water and ice were mixed in a copper calorimeter - Leaving Cert Physics - Question 3 - 2017

To ensure that the temperature of all the added ice was at 0.0 °C, the ice could be:

• Crushed into smaller pieces, allowing it to reach thermal equilibrium more quickly.
• Allowed to melt completely in a controlled environment, ensuring that it maintains the melting point.

1. A named insulator could be wrapped around the calorimeter to minimize heat exchange with the environment.
2. A lid could be placed over the calorimeter to further reduce heat loss.

The specific latent heat of fusion of ice can be calculated using the principle of conservation of energy. The heat lost by the warm water and calorimeter is equal to the heat gained by the melting ice:

Let:

• $m_w = 48.4 \text{ g}$ (mass of warm water)
• $c_w = 4180 \text{ J kg}^{-1} \text{ K}^{-1}$ (specific heat capacity of water)
• $\Delta T_w = 12.0 - 26.5$ (change in temperature of warm water)

Heat lost by warm water: $Q_{lost} = m_w c_w \Delta T_w$

Also, let:

• $m_{ice} = 8.2 \text{ g}$ (mass of added ice)
• $L =$ specific latent heat of fusion of ice

Heat gained by melting ice: $Q_{gained} = m_{ice} L$

Setting these equal gives: $Q_{lost} = Q_{gained}$

Substituting the values leads to: (\text{(61.8 + m_{ice})} \cdot c_{water} \cdot (12.0 - T_{final})) = m_{ice}L

After performing the calculations, the specific latent heat of fusion of ice is found to be: $L = 3.5 \times 10^5 \text{ J kg}^{-1}$

1. It should have a small heat capacity, minimizing the energy absorbed by the thermometer itself.
2. It should be graduated to 0.1 °C for precise measurements and able to react quickly to changes in temperature.