Question

A copper block of mass 2.5 kg is heated in a furnace to a temperature of $500°\mathrm{C}$ and then placed on a large ice block. What is the maximum amount of ice that can melt? (specific heat of copper= $0.39\mathrm{j}{\mathrm{g}}^{-1}{\mathrm{k}}^{-1}$ ; heat of fusion of water = $335\mathrm{j}{\mathrm{g}}^{-1}$)

Answer 1

Step 1: Given data

Mass of copper = $2.5\mathrm{kg}$

Rise is temperature = $500°\mathrm{C}$

Specific heat of copper= $0.39\mathrm{j}{\mathrm{g}}^{-1}{\mathrm{k}}^{-1}$

The heat of fusion of water = $335\mathrm{j}{\mathrm{g}}^{-1}$

Step 2: Calculating heat lost by copper

$\therefore$The amount of heat lost by the copper block is

$$\begin{gathered} \mathrm{Q}=\mathrm{mC}∆\mathrm{T} \\ \Rightarrow \mathrm{Q}=2.5\times 1000\times .39\times 500 \\ \Rightarrow \mathrm{Q}=2500\times 195 \\ \Rightarrow \mathrm{Q}=487500\mathrm{J} \end{gathered}$$

Step 3: Calculating amount of ice that can melt

Let us assume that the amount of ice that can melt be ${\mathrm{m}}_1\mathrm{g}$when the copper is placed on the ice block

$\therefore$The amount of heat gained by the melted ice is $\mathrm{Q}={\mathrm{m}}_1\mathrm{L}$

$$\begin{gathered} \therefore \mathrm{Q}={\mathrm{m}}_1\mathrm{L} \\ \Rightarrow {\mathrm{m}}_1=\frac{\mathrm{Q}}{\mathrm{L}} \\ \Rightarrow {\mathrm{m}}_1=\frac{487500}{335} \\ \Rightarrow {\mathrm{m}}_1=1455.223\mathrm{g} \\ \Rightarrow {\mathrm{m}}_1=1.45\mathrm{kg} \end{gathered}$$

Therefore, the maximum amount of ice that can melt is $1.45\mathrm{kg}$.