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Question

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 (approx.) of ice that can melt? (Specific heat of copper sc=0.39 J/gC, latent heat of fusion of water L=335 J/g).

A
0.5 kg
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B
1.5 kg
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C
2.5 kg
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D
2 kg
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Solution

The correct option is B 1.5 kg
Formula used: Q=mL,Q=msΔ θ

Given: mc=2.5 kg, θ1=0 C, θ2=500 C, sc=0.39 J/gC, L=335J/g

Fall in temperature of the copper block,
Δ T=500 C0 C=500 C
If m is the mass of ice melted, applying the principle of calorimetry,
Heat gained by ice=heat lost by copper block
i.e., mL=mcsc Δ T
or, m=mc sc Δ TL
m=2.5×103×0.39×500335
m=1455 g=1.455kg1.5 kg

FINAL ANSWER: (b).

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