A copper block of mass  $2.5kg$ is heated in a furnace to a temperature of  ${500}^{\circ }\mathrm{C}$ and then placed on a large ice block. What in the maximum amount of ice that can melt in kg? (Specific heat of copper $=0.39{\mathrm{Jg}}^{-1}{\mathrm{K}}^{-1};$  heat of fusion of water $=335{\mathrm{Jg}}^{-1}$) .

Specific heat of copper,  $c=0.39{\mathrm{Jg}}^{-1}{K}^{-1}$

$\Rightarrow c=0.39\times {10}^3{\mathrm{Jkg}}^{-1}{\mathrm{K}}^{-1}$
  
Temperature of furnace, $\mathrm{\Delta }\theta ={500}^{\circ }C$ 
Latent heat of fusion,  $L=335{\mathrm{Jg}}^{-1}$

$\Rightarrow L=335\times {10}^3{\mathrm{Jkg}}^{-1}$
  
If Q be the heat absorbed by the copper block, then 

$Q=m_{1}c\mathrm{\Delta }\theta$......…(i)
Let $m_2$  be the mass of ice melted, when copper block is placed on it, then
 $\therefore Q=m_{2}L$...........(ii)
     From Eqs. (i) and (ii) we get
  $\begin{array}{l}{m}_{1}c\mathrm{\Delta }\theta =m_{2}L\\ m_2=\frac{m_{1}c\mathrm{\Delta }\theta }{L}=\frac{2.5\times 0.39\times {10}^3\times 500}{335\times {10}^3}\\ =1.455\mathrm{kg}\\ =1.46\mathrm{kg}\end{array}$