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A copper block of mass $2.5\; kg$ 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 of ice (in $kg$) that can melt? (Specific heat of copper $=0.39\; J g ^{-1} K ^{-1}$ ;heat of fusion of water $=335\; J g ^{-1})$
$1.45$
$1$
$2.1$
$2.5$
Solution
Mass of the copper block, $m=2.5 kg =2500 g$
Rise in the temperature of the copper block, $\Delta \theta=500^{\circ} C$
Specific heat of copper, $C=0.39 Jg ^{-1\;\circ} C ^{-1}$
Heat of fusion of water, $L=335 Jg ^{-1}$
The maximum heat the copper block can lose, $Q=m C \Delta \theta$
$=2500 \times 0.39 \times 500$
$=487500 J$
Let $m_{1} g$ be the amount of ice that melts when the copper block is placed on the ice block.
The heat gained by the melted ice, $Q=m_{1} L$ $\therefore m_{1}=\frac{Q}{L}=\frac{487500}{335}=1455.22 g$
Hence, the maximum amount of ice that can melt is $1.45 \,kg$.
