A metal block of mass $3.3 \mathrm{~kg}$ is heated to a temperature of $400^{\circ} \mathrm{C}$ and then…
A metal block of mass $3.3 \mathrm{~kg}$ is heated to a temperature of $400^{\circ} \mathrm{C}$ and then block is placed on large ice block. The specific heat of the metal is $0.4 \mathrm{~J} \mathrm{~g}^{-1} \mathrm{~K}^{-1}$ and heat of fusion of water is $330 \mathrm{Jg}^{-1}$. The maximum amount of ice that can melt is
$1.2 \mathrm{~kg}$
$2.2 \mathrm{~kg}$
$1.6 \mathrm{~kg}$
$2 \mathrm{~kg}$
Solution
Given, mass of metal block, $m=33 \mathrm{~kg}$
Temperature, $\Delta T=400^{\circ} \mathrm{C}$
Specific heat of metal block, $s_m=0.4 \mathrm{Jg}^{-1} \mathrm{~K}^{-1}$
$
=400 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}
$
Latent heat of fusion, $L_f=330 \mathrm{Jg}^{-1}=3.3 \times 10^5 \mathrm{Jkg}^{-1}$
If $m^{\prime}$ be the maximum amount of mass of melted ice, then
$
\begin{aligned}
m^{\prime} L_f & =m s_m \Delta T \\
\Rightarrow \quad m^{\prime} & =\frac{m s_m \Delta T}{L_f} \\
& =\frac{3.3 \times 400 \times 400}{3.3 \times 10^5}=1.6 \mathrm{~kg}
\end{aligned}
$