Question:
A metal block of mass \(3.3\,\text{kg}\) is heated to a temperature of \(400^\circ\text{C}\) and then block is placed on a large ice block. The specific heat of the metal is \(0.4\,\text{J g}^{-1}\text{K}^{-1}\) and heat of fusion of water is \(330\,\text{J g}^{-1}\). The maximum amount of ice that can melt is
A metal block of mass \(3.3\,\text{kg}\) is heated to a temperature of \(400^\circ\text{C}\) and then block is placed on a large ice block. The specific heat of the metal is \(0.4\,\text{J g}^{-1}\text{K}^{-1}\) and heat of fusion of water is \(330\,\text{J g}^{-1}\). The maximum amount of ice that can melt is
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For ice-melting problems, equate heat lost by the hot body to latent heat gained by ice:
\[
mc\Delta T=m_{\text{ice}}L
\]
Updated On: Jun 26, 2026
- \(1.2\,\text{kg}\)
- \(2.2\,\text{kg}\)
- \(1.6\,\text{kg}\)
- \(2\,\text{kg}\)
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The Correct Option is C
Solution and Explanation
Step 1: Convert the mass of metal into grams.
Given mass of metal: \[ 3.3\,\text{kg}=3300\,\text{g} \] Specific heat of metal: \[ c=0.4\,\text{J g}^{-1}\text{K}^{-1} \] Initial temperature: \[ 400^\circ\text{C} \] Final temperature will be \[ 0^\circ\text{C} \] because the metal is placed on ice.
So, \[ \Delta T=400 \]
Step 2: Calculate heat lost by the metal.
Heat lost is \[ Q=mc\Delta T \] \[ Q=3300\times 0.4\times 400 \] \[ Q=528000\,\text{J} \]
Step 3: Use latent heat of fusion of ice.
Heat required to melt ice is \[ Q=mL \] Given, \[ L=330\,\text{J g}^{-1} \] Let the mass of ice melted be \(m\) grams.
Then, \[ m=\frac{Q}{L} \] \[ m=\frac{528000}{330} \] \[ m=1600\,\text{g} \] \[ m=1.6\,\text{kg} \]
Step 4: Final conclusion.
Hence, the maximum amount of ice that can melt is \[ \boxed{1.6\,\text{kg}} \]
Given mass of metal: \[ 3.3\,\text{kg}=3300\,\text{g} \] Specific heat of metal: \[ c=0.4\,\text{J g}^{-1}\text{K}^{-1} \] Initial temperature: \[ 400^\circ\text{C} \] Final temperature will be \[ 0^\circ\text{C} \] because the metal is placed on ice.
So, \[ \Delta T=400 \]
Step 2: Calculate heat lost by the metal.
Heat lost is \[ Q=mc\Delta T \] \[ Q=3300\times 0.4\times 400 \] \[ Q=528000\,\text{J} \]
Step 3: Use latent heat of fusion of ice.
Heat required to melt ice is \[ Q=mL \] Given, \[ L=330\,\text{J g}^{-1} \] Let the mass of ice melted be \(m\) grams.
Then, \[ m=\frac{Q}{L} \] \[ m=\frac{528000}{330} \] \[ m=1600\,\text{g} \] \[ m=1.6\,\text{kg} \]
Step 4: Final conclusion.
Hence, the maximum amount of ice that can melt is \[ \boxed{1.6\,\text{kg}} \]
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