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A block of mass $2.5\,\, kg$ is heated to temperature of $500^o C$ and placed on a large ice block. ......... $kg$ is the maximum amount of ice that can melt (approx.). Specific heat for the body $= 0.1 Cal/gm^o C$.
$1$
$1.5$
$2 $
$2.5$
Solution
Here, mass of copper block,
$M=2.5 \mathrm{kg}=2.5 \times 10^{3} \mathrm{g}$
Rise in temperature of the copper block,
$\Delta T=500^{\circ} \mathrm{C}-0^{\circ} \mathrm{C}=500^{\circ} \mathrm{C}$
Specific heat of copper, $c=0.39 J / g^{\circ} C$
Lalent heat of fusion of water, $L=335 J / g$
if $m$ is the mass of ice melted, applying the principle of calorimetry.
Heat gained by ice=heat lost by copper block
i.e., $m L=M c \Delta T$
or, $m=\frac{M c \Delta T}{L}$
$=\frac{2.5 \times 10^{3}(g) \times 0.39\left(J / g C^{0}\right) \times 500\left(C^{\circ}\right)}{335(J / g)}$
$=1455 g=1.455 \approx 1.5 k g$
