In an experiment to measure the specific latent heat of fusion of ice, warm water was placed in a copper calorimeter - Leaving Cert Physics - Question 2 - 2008

Dried, melting ice was used because it removes any pre-existing water. Melted ice would have already absorbed latent heat, which could skew the results. Thus, only ice that has not absorbed heat from the environment is added.

The mass of the ice was calculated by subtracting the initial mass of the calorimeter and water from the total mass recorded after adding the ice. This can be expressed as:

final mass of calorimeter + contents - mass of calorimeter + water = mass of ice.

The approximate room temperature should be around 20 ± 1.0 °C. This is halfway between the initial and final temperatures of the water in the calorimeter.

The energy lost is calculated as:

$ext{Energy lost} = (mC_{ ext{water}} riangle T) + (mC_{ ext{cal}} riangle T)$

where:

• $m$ is the mass,
• $C$ is the specific heat capacity,
• $riangle T$ is the change in temperature.

Applying the values:

$= (0.0605 imes 390 imes (30.5-10.2)) + (0.0583 imes 4200 imes (30.5-10.2))$ $= 5494.6365 ext{ J}$

The specific latent heat of fusion can be calculated using:

$L = \frac{Q}{m}$

where:

• $Q$ is the energy gained by ice and melted ice,
• $m$ is the mass of ice.

Calculating:

$L = \frac{(mC_{w} (T_f - T_i))}{m_{ice}} = \frac{(0.0151)(4200)(10.2)}{0.0151} + 646.984 = 5494.6365 ext{ J kg}^{-1}$