Question

A $2\, kg$ copper block is heated to $500^{\circ} C$ and then it is placed on a large block of ice at $0^{\circ} C$. If the specific heat capacity of copper is $400\, J - kg ^{-1}{ }^{\circ} C ^{-1}$ and latent heat of fusion of water is $3.5 \times 10^{5} J - kg ^{-1}$, the amount of ice that can melt is

• A. $\frac{7}{8} kg$
• B. $\frac{7}{5} kg$
• C. $\frac{8}{7} kg$
• D. $\frac{5}{7} kg$

Answer 1

Answer: (C)

Heat emitted by copper = Heat gained by ice
$mc \Delta \theta= m ^{\prime} L$
$\Rightarrow m '=\frac{ mc \Delta \theta}{ L }$
Given, $m=2\, kg , c=400\, J - kg ^{-1} C ^{-1},$
$\Delta \theta=500, L=3.5 \times 10^{5} J - kg ^{-1}$
$\therefore m'=\frac{2 \times 400 \times 500}{3.5 \times 10^{5}}=\frac{8}{7} kg$