Question

A copper block of mass $2.5kg$ 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 $s_c=0.39J/g^{\circ }C$, latent heat of fusion of water $L=335J/g$).

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

Answer 1

The correct option is B $1.5kg$

Formula used: $Q=mL,Q=ms\Delta \theta$

Given: $m_c=2.5kg,{\theta }_1=0^{\circ }C,{\theta }_2={500}^{\circ }C,s_c=0.39J/g^{\circ }C,L=335J/g$

Fall in temperature of the copper block,
$\Delta T={500}^{\circ }C-0^{\circ }C={500}^{\circ }C$
If $m$ is the mass of ice melted, applying the principle of calorimetry,
Heat gained by ice=heat lost by copper block
i.e., $mL=m_c{s}_c\Delta T$
or, $m=\frac{m_c{s}_c\Delta T}{L}$
$m=\frac{2.5\times {10}^3\times 0.39\times 500}{335}$
$m=1455g=1.455kg\approx 1.5kg$

FINAL ANSWER: (b).