Q. 9 C3.7( 36 Votes )

Solve the followi

Answer :

1 g of ice needs 80 cal to become water at 0 C.
1 g of water needs 50 cal to rise to 50 C from 0 C.
So total the heat needed is 150×80 + 150×50 = 19500 cal.
1 g of steam will supply 540 c to become water at 100 C.
1 g of water at 100 C will supply 50 cal to become water at 50 C.
Hence x gram of steam at 100 C will supply 590 cal while cooling to 50 C.
Hence 19500 = 590 x.
Solving, x = 33.05 gram. 

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