Steam at100^(@)C is passed into 20g...

Steam at`100^(@)C ` is passed into 20g of water at `10^(C` when water acquires temp of `80^(@)C` the mass of water present will be (take specifies heat of water 1 cal `g^(1) c^(-1)` & latent heat of steam =540 cal `g^(-1) c^(-1)`

`42.5 g`

`22.5 g`

`24 g`

`31 . 5g`

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Steam at 100^@ C is passed.into 20 g of water at 10^@C Whenwater acquires a temperature of 80^@C the mass of water present will be :[take specific heat of steam =1 cal g^-1^@C^-1 and latent of water =540 cal g^-1 ]

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