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 to avoid any additional water from melted ice. If melted ice were used, it would have already gained latent heat, thus affecting the measurement. Using dried ice ensures that no water is added, and it starts melting at 0 °C, which stabilizes the experimental conditions.

The mass of the ice was determined by finding the final mass of the calorimeter combined with the contents of the calorimeter, and then subtracting the mass of the calorimeter and water.

The approximate room temperature should be around 20 ± 1 °C. This temperature is midway between the initial and final temperatures of the water in the calorimeter.

The energy lost can be calculated using the formula:

$E = mc heta$

For the calorimeter, the energy lost is:

$E_{cal} = (m_{cal} imes c_{c} imes heta_{cal})$

For the water, the energy lost is:

$E_{water} = (m_{water} imes c_{water} imes heta_{water})$

Substituting the values:

$E_{lost} = (0.0605 imes 390 imes (30.5 - 10.2)) + (0.0583 imes 4200 imes (30.5 - 10.2))$

Calculating gives:

$E_{lost} = 5449.6365 ext{ J}$

The specific latent heat of fusion of ice can be calculated using the energy gained by the ice as it melts:

$L = \frac{E_{gain}}{m_{ice}}$

Here, the energy gained by ice can be written as:

$E_{gain} = m_{ice} imes L + m_{ice} imes c_{water} imes (10.2 - 0)$

Substituting the values:

$L = \frac{(0.0151 imes 4200 imes 10.2)}{0.0151} + 646.844$

Calculating gives:

$L = 3.18 imes 10^5 ext{ J kg}^{-1}$