Steam at `100^(@)C` is passed into `20 g` of water at `10^(@)C` when water acquire 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. 42.5 g
B. 22.5 g
C. 24 g
D. 31.5 g

Answer

B

Explanation

Heat lost by steam = heat gained by water
`m.L + m. S Delta T = m S Delta T`
`rArr m. xx 540 + m. xx 1 xx (100 - 80) = 20 xx 1 (80-10)`
`rArr m. = (20 xx 70)/(560) = 2.5g`
Net mass of water = 20 + 2.5 g =22.5g.