Stream at `100^(@)C` is passed into 20 g of water at `10^(@)C`. When water acquires a temperature of `80^(@)C`, the mass of water present will be [Take specific heat of water `=1 cal g^(-1) .^(@)C^(-1)` and latent heat of steam `=540 cal g^(-1)`]

Options

A. 24 g
B. 31.5g
C. 42.5g
D. 22.5 g

Answer

D

Explanation

Heat lost by system = Heat gain by water.
Let m' be amount of heat converted into water.
`m^(') xx 540 = 20 xx1xx(80-10)`
`m^(')= (20 xx70)/(540)` = 2.5g
Now, net mass of water = 20+2.5 = 22.5g