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