In an experiment to determine the specific latent heat of fusion of ice, a student first crushed some ice - Leaving Cert Physics - Question 3 - 2019

Drying the ice ensures that only ice enters the calorimeter, which allows for accurate measurement of the latent heat of fusion. If water were present, it would absorb energy, skewing the results.

The ice was crushed using an ice crusher, a tool designed specifically for breaking ice into smaller pieces.

The ice was dried using a towel, which removes any surface moisture and prevents additional water from entering the calorimeter.

Using warm water helps to ensure that energy is gained and transferred effectively. This reduces the time taken for the ice to melt and stabilizes the temperature in the calorimeter.

Melting ice was used because it provides a constant temperature of 0 °C, allowing calculations for the latent heat of fusion to be more accurate.

To calculate the specific latent heat of fusion of ice, we first determine the mass of ice:

Mass of water before adding ice = 108.5 g - 56.3 g = 52.2 g.

Now, we calculate the heat lost by warm water:

$\Delta \theta_0 = T_0 - T_f = 29.5 °C - 8.0 °C = 21.5 °C$

Using the formula $Q = mc\Delta \theta$, we find:

$Q = (52.2 g)(4180 kg^{-1} k^{-1})(21.5)$

Calculate: $Q \approx 4.63 \times 10^3 J$

The heat gained by ice is:

$Q = mL$

Using the mass of ice, which is:

$m = 14.0 g$

Setting these two equations equal gives:

$4.63 \times 10^3 J = (14.0 g)L$

Solving for L gives:

$L = \frac{4.63 \times 10^3 J}{14.0 g} \approx 3.25 \times 10^5 kg^{-1}$

Using a very large mass of water could lead to smaller changes in temperature and greater percentage errors. A larger mass results in a smaller temperature change, making it difficult to accurately gauge the heat transfer, which may introduce errors in calculating the specific latent heat.