Question

`50 g` of steam at `100^@ C` is passed into `250 g` of at `0^@ C`. Find the resultant temperature (if latent heat of steam is `540 cal//g`, latent heat of ice is `80 cal//g` and specific heat of water is `1 cal//g -.^@ C`).
A. `40^@ C`
B. `30^@ C`
C. `20^@ C`
D. `10^@ C`

Answer 1

Correct Answer - A
Heat lost by steam = Heat gained by ice.
`m_(steam) L_(v)+m_(s) xxS_(w)(100^@-theta^@)`
`=m_(w)L_(f) +m_(w)S_(w) (theta^@ - 0)`.