Question

# $$60 g$$ of ice at $$0^0$$ is mixed with $$60 g$$ of steam at $$100^0$$. At thermal equilibrium, the mixture contains:(Latent heats of steam and ice are $$540 cal { g }^{ -1 }$$ and $$80 cal { g }^{ -1 }$$ respectively, specific heat of water $$= 1 cal { g }^{ -1 } { }^{ -1 }$$)

A
80g of water and 40g of steam at 1000
B
120g of water at 900
C
120g of water at 1000
D
40g of steam and 80g of water at 00

Solution

## The correct option is A $$80 g$$ of water and $$40 g$$ of steam at $$100 ^0$$Heat energy required to melt $$60 g$$ of ice $$= 60\times 80=4800 Cal$$.Heat energy required to raise the temprature of water at $$0$$ degree centigrade to $$100$$ degree centigrade $$= 100\times 1\times 60=6000 Cal$$Total energy required $$= 4800+6000=10800 Cal$$This energy is given by the steam by getting converted into water.Amount of steam getting converted into water is $$\dfrac{10800}{540}=20g$$Thus, finally total $$80 gm$$ of water will be there and $$40 gm$$ of steam. Physics

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