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A copper block of mass $2.5 \mathrm{~kg}$ is heated in a furnace to a temperature of $500^{\circ} \mathrm{C}$ and then placed on a large ice block. What is the maximum amount of ice that can melt?
Specific heat of copper is $=0.39 \mathrm{Jg}^{-1} \mathrm{~K}^{-1}$, heat of fusion of water $=\mathbf{3 3 5} \mathrm{Jg}^{-1}$
Specific heat of copper is $=0.39 \mathrm{Jg}^{-1} \mathrm{~K}^{-1}$, heat of fusion of water $=\mathbf{3 3 5} \mathrm{Jg}^{-1}$
Solution:
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Verified Answer
Mass of copper $=m_1=2.5 \mathrm{~kg}=2500 \mathrm{~g}$,
Change in temperature $=\Delta t=500-0=500^{\circ} \mathrm{C}$
Specific heat of copper $=c_1=0.39 \mathrm{~J} / \mathrm{g} /{ }^{\circ} \mathrm{C}$,
Latent heat of fusion of ice $=\mathrm{L}=335 \mathrm{~J} / \mathrm{g}$
Let mass of ice melted $=m_2$
$\because$ Heat lost by copper $=$ Heat gained by ice
$\Rightarrow m_1 c_1 \Delta t=\mathrm{m}_2 \mathrm{~L}$
$m_2=\frac{m_1 c_1 \Delta t}{\mathrm{~L}}=\frac{2500 \times 0.39 \times 500}{335}$
$=0.1455 \times 10^4=1455 \mathrm{~g}$
Change in temperature $=\Delta t=500-0=500^{\circ} \mathrm{C}$
Specific heat of copper $=c_1=0.39 \mathrm{~J} / \mathrm{g} /{ }^{\circ} \mathrm{C}$,
Latent heat of fusion of ice $=\mathrm{L}=335 \mathrm{~J} / \mathrm{g}$
Let mass of ice melted $=m_2$
$\because$ Heat lost by copper $=$ Heat gained by ice
$\Rightarrow m_1 c_1 \Delta t=\mathrm{m}_2 \mathrm{~L}$
$m_2=\frac{m_1 c_1 \Delta t}{\mathrm{~L}}=\frac{2500 \times 0.39 \times 500}{335}$
$=0.1455 \times 10^4=1455 \mathrm{~g}$
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