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One gram of ice at 0^(@)C is added to 5 ...

One gram of ice at `0^(@)C` is added to 5 gram of water at `10^(@)C`. If the latent heat of ice be 80 cal/g, then the final temperature of the mixture is -

A

`(5M_1)/(M_2)-50`

B

`(50M_2)/(M_1)-5`

C

`(50 M_2)/(M_1)`

D

`(5M_2)/(M_1)-5`

Text Solution

Verified by Experts

The correct Answer is:
B
Heat taken by ice = Heat given by water
`M_1S_("ice")(10)+M_1L_f=M_2S_W(50)`
`(M_1)/(2)xx10+M_1L_f=M_2xx50`
`5+L_f=50(M_2)/(M_1)`
`L_f=(50M_2)/(M_1)-5`
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