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Q.

The specific heat of water = 4200 Jkg−1K−1  and the latent heat of ice = 3.4×105Jkg−1. 100g of ice at 0°C is placed in 200g of water at 25°C . The amount of ice that will melt as the temperature of water reaches 0°C  is close to (in grams)

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a

64.6

b

63.8

c

69.3

d

61.7

answer is D.

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Detailed Solution

Let the mass of ice which gets melted be m1 , Now
Heat lost by water = Heat gained by ice                          ⇒mwcwΔTw=miLi    
 ⇒2001000×4200×25−0=mi×3.4×105        ⇒21000=mi×3.4×105

⇒mi=210003.4×105=0.0617kg=61.7g

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