Question

How much heat is necessary to change 460 g of ice at -18°C to water at 20°C? kcal

Answer 1

Given:

The mass of ice is,

`m=460\text{ kg}`

The initial temperature of ice is,

`t_i=-18^(\circ)C`

The heat for this ice to become 0 degree C ice is,

`\begin{gathered} H_1=460*0.5*18 \\ =4.14\text{ kcal} \end{gathered}`

The heat needed for 0 degree ice to 0 degree water is,

`\begin{gathered} H_2=460*80 \\ =36.8\text{ kcal} \end{gathered}`

Now the heat for the final step to reach 20 degrees is,

`\begin{gathered} H_3=460*1*(20-0) \\ =9.2\text{ kcal} \end{gathered}`

The total heat is,

`\begin{gathered} H=4.14+36.8+9.2 \\ =50.1\text{ kcal} \end{gathered}`

Hence the total heat is 50.1 kcal.