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Q.

Steam at 100 °C is mixed with 10 g of ice at 0 °C so as to get water at a temperature of 40 °C. The total mass of water at 40 °C is
(Latent heat of steam = 540 cal g-1; Latent heat of fusion of ice = 80 cal g-1; Specific heat of water =1cal g-1 C-1)

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a

12.22g

b

12g

c

10g 

d

2g 

answer is C.

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Detailed Solution

x×540+x×(100-40)=10×80 + 10×1×(40-0)

On solving for x, 

x = 2 g

Total amount of water at 400 is 12 g.

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