Steam at `100^(@)C` is passed into 20 g of water at `10^(@)`C, then water acquires a temperating of `80^(@)`C, the man 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. 31.5 g
C. 42.5 g
D. 22.5 g

Explanation

(d) Heat gain by water = heat loss by steam
`20 xx 1 xx (80 –100) = m xx 540 + m xx 1 × (100 – 80)`
1400=560 m
`m= (1400)/(560)` =2.5 g
Total mass of water = 20 + 2.5 = 22.5 g