A copper block of mass  $5.0kg$ is heated to a temperature of  $500°C$ and is placed on a large ice block. What is the maximum amount of ice that can melt? [Specific heat of copper = $0.39J{g}^{-1}°C^{-1}$, and latent heat of fusion of water =$335J{g}^{-1}$ ]

The mass of copper block, $m=5kg$ , Initial temperature of copper block, ${T_1}_{}=500°C$ 
Final temperature of copper block, when it is placed on the block, ${T_2}_{}=0°C$  
Specific heat of copper, $s=0.39\times J{g}^{-1}°C^{-1}=0.39\times {10}^{3}JK{g}^{-1}°C^{-1}$  
$\therefore$ Heat given by the block, $\Delta Q=ms\Delta T$ ,  $\Delta Q=5\times 0.39\times {10}^3\times \left(500-0\right)=9.75\times {10}^{5}J$
Heat taken by ice to melt m kg of ice ,  $Q_2=m{L}_f=m\times 335\times {10}^3$  

$\left(\because L_f=335J{g}^{-1}=335\times {10}^{3}JK{g}^{-1}\right)$
Hence, according to principle of calorimetry, $Q_2=Q_1$ 

$\mathrm{m}\times 335\times {10}^3=9.75\times {10}^5\Rightarrow \mathrm{m}=\frac{9.75\times {10}^5}{335\times {10}^3}=2.91\mathrm{kg}$