Question

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)`]
A. `24 g`
B. `31.5 g`
C. `42.5 g`
D. `22.5 g`

Answer 1

Correct Answer - D
Heat gain by water `=` Heat lost by steam
`m_(w)s_(w) Delta theta_(1) = mLv + ms_(w)Delta theta_(2)`
`20xx1 xx (80-10)=mxx540+mxx1xx(100-80)`
`rArr 1400 = 560m rArr m = 2.5 g`
Total mass of water `= 20+2.5 = 22.5g`.