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m(1) gram of ice at -10^(@)C and m(2) gr...

`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 :

A

`(50M_(2))/(M_(1))`

B

`(5M_(2))/(M_(1))-5`

C

`(5M_(1))/(M_(2))-50`

D

`(50M_(2))/(M_(1))-5`

Text Solution

Verified by Experts

The correct Answer is:
D
`Q_("gain")=Q_("lost")`
`M_(1)xx0.5xx10+M_(2)xx1xx50," "L=(M_(2))/(M_(1))xx50-5`
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