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 was crucial to ensure that only ice was added to the calorimeter, preventing water from the melting ice from affecting the results.

The ice was crushed using an ice crusher, which breaks the ice into smaller pieces to enhance the melting process.

The ice was dried using a towel to remove any moisture from its surface before it was introduced into the calorimeter.

Warm water was used because it would melt the ice more quickly and efficiently, leading to a more accurate measurement of energy transfer.

Using melting ice ensures that the ice is at 0 °C, which is essential for accurately calculating the latent heat of fusion.

Using the formula $ext{Q} = mL$ where $Q$ is the heat absorbed, $m$ is the mass of the ice, and $L$ is the specific latent heat of fusion.

The initial temperature change: $heta_0 = 8.0 °C - 29.5 °C = -21.5 °C$

The mass of water: $m = 122.9 g - 108.5 g = 14.4 g = 0.0144 kg$

Now, calculate energy gained from the water: $ext{Energy from water} = m imes c imes heta$ $ext{Energy from water} = 0.0144 imes 4180 imes 21.5 = 1294.08 J$

Now, equating this to the energy absorbed by the ice: $Q_{ice} = ext{mass of ice} imes L$

Assuming all temperature change is accounted for: L = rac{1294.08 J}{ ext{mass of ice}} = ext{mass of ice} imes L

Further calculations will provide the value of $L$.

Using a large mass of water could result in a smaller change in temperature, leading to greater percentage errors in calculating specific heat, which can compromise the overall accuracy of the experiment.