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Q.

The specific heat of water = $4200 {Jkg}^{- 1} K^{- 1}$ and the latent heat of ice = $3.4 \times 10^{5} J k g^{- 1}$ . 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 $m_{1}$ , Now
Heat lost by water = Heat gained by ice $\Rightarrow m_{w} c_{w} Δ T_{w} = m_{i} L_{i}$
$\Rightarrow \frac{200}{1000} \times 4200 \times (25 - 0) = m_{i} \times 3.4 \times 10^{5}$ $\Rightarrow 21000 = m_{i} \times 3.4 \times 10^{5}$

$\Rightarrow m_{i} = \frac{21000}{3.4 \times 10^{5}} = 0.0617 k g = 61.7 g$

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