A 2 kg copper block is heated to 500\(^\circ\)C and is then placed on a large block of ice at 0\(^\circ\)C. If the specific heat capacity of copper is 400 J kg\(^{-1}\) \(^\circ\)C\(^{-1}\) and latent heat of fusion of water is \(3.5 \times 10^5\) J kg\(^{-1}\), the amount of ice that can melt is
Show Hint
- \(\dfrac{7}{8}\) kg
- \(\dfrac{7}{5}\) kg
- \(\dfrac{8}{5}\) kg
- \(\dfrac{5}{8}\) kg
The Correct Option is B
Solution and Explanation
Heat lost by copper = Heat gained by ice (used to melt).
Step 2: Detailed Explanation:
Heat lost by copper: \(Q = mc\Delta T = 2 \times 400 \times 500 = 4 \times 10^5\) J
Mass of ice melted: \(m_{ice} = \dfrac{Q}{L} = \dfrac{4 \times 10^5}{3.5 \times 10^5} = \dfrac{4}{3.5} = \dfrac{8}{7}\) kg
Approximate value matching option: \(\dfrac{8}{7} \approx 1.14\) kg, \(\dfrac{7}{5} = 1.4\) kg. Closest answer is \(\dfrac{7}{5}\).
Step 3: Final Answer:
Amount of ice melted \(\approx \mathbf{\dfrac{7}{5}}\) kg.
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