Question

A copper block of mass $2.5\, kg$ is heated in a furnace to a temperature of $500^{\circ} C$ and then placed on a large ice block. What is the maximum amount (approx.) of ice that can melt? (Specific heat of copper $=0.39\, J / g ^{\circ} C$ heat of fusion of water $=335\, J / g$ ).

• A. 2 kg
• B. 1.5 kg
• C. 2.5 kg
• D. 0.5 kg

Answer 1

Answer: (B)

Here, mass of copper block,
$M=2.5\, kg =2.5 \times 10^{3} g$
Rise in temperature of the copper block,
$\Delta T=500^{\circ} C -0^{\circ} C =500^{\circ} C$
Specific heat of copper, $c=0.39\, J / g ^{\circ} C$
Latent heat of fusion of water, $L=335\, J / g$
If $m$ is the mass of ice melted, applying the principle of calorimetry,
Heat gained by ice $=$ heat lost by copper block
i.e., $m L=M c \Delta T$
or $m=\frac{M c \Delta T}{L}$
$=\frac{2.5 \times 10^{3}(g) \times 0.39\left( J / gC ^{\circ}\right) \times 500\left( C ^{\circ}\right)}{335( J / g )}$
$=1455\, g =1.455 \approx 1.5\, kg$