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Thermal Properties of Matter

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Question : 13 of 22
Marks: +1, -0
A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500°C and then placed on a large ice block. What is the maximum amount of ice that can melt? (Specific heat of copper =0.39 J g−1 K−1= 0.39\,\mathrm{J}\,\mathrm{g}^{-1}\,\mathrm{K}^{-1}; heat of fusion of water =335 J g−1)= 335\,\mathrm{J}\,\mathrm{g}^{-1}).
Solution:  
Here, mass of copper block, m1=2.5 kgm_{1}=2.5\,\mathrm{kg}
Specific heat of copper,
C=0.39 J g−1 K−1C=0.39\,\mathrm{J}\,\mathrm{g}^{-1}\,\mathrm{K}^{-1}
=0.39×103 J kg−1 K−1=0.39\times 10^{3}\,\mathrm{J}\,\mathrm{kg}^{-1}\,\mathrm{K}^{-1}
Temperature of furnace, ΔT=500∘C\Delta T = 500^{\circ}\mathrm{C}
Latent heat of fusion,
L=335 J g−1L=335\,\mathrm{J}\,\mathrm{g}^{-1}
=335×103 J kg−1=335\times 10^{3}\,\mathrm{J}\,\mathrm{kg}^{-1}
If QQ be the heat absorbed by the copper block, then
Q=m1CΔTQ=m_{1} C \Delta T... (i)
Let m2(kg)m_2 (\mathrm{kg}) be the mass of ice melted when copper block is placed on it,∴Q=m2L\therefore Q = m_2 L... (ii)
∴ from (i) and (ii), we get, m1CΔT=m2Lm_1 C \Delta T = m_2 L
m2=m1CΔTLm_{2}= \frac{ m_{1} C \Delta T }{ L }
or 2.5×0.39×103×500335×103\frac{2.5 \times 0.39 \times 10^{3} \times 500}{335 \times 10^{3}}
=1.155 kg=1.5 kg=1.155\,\mathrm{kg}=1.5\,\mathrm{kg}.
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