`m_(1)` gram of ice at `-10^(@)C` and `m_(2)` gram of water at `50^(@)C` are mixed in insulated container. If in equilibrium state we get only water at `0^(@)C` then latent heat of ice is :

Options

A. `(50M_(2))/(M_(1))`
B. `(5M_(2))/(M_(1))-5`
C. `(5M_(1))/(M_(2))-50`
D. `(50M_(2))/(M_(1))-5`

Answer

D

Explanation

`Q_("gain")=Q_("lost")`
`M_(1)xx0.5xx10+M_(1)L=M_(2)xx1xx50`
`L=(M_(2))/(M_(1))xx50-5`