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M' kg of water 't' 0^@ C is divided into...

M' kg of water 't' `0^@ C` is divided into two parts so that one part of mass 'm' kg when converted into ice at `0^@ C` would release enough heat to vapourise the other part, then `(m)/(M)` is equal to [Specific heta of water `= 1 cal g^(-1) .^@ C^(-1)`,
Latent heat of fusion of ice `= 80 cal g^(-1)`,
Latent heat of steam `= 540 cal g^(-1)`].

A

`640-t`

B

`(720-t)/(640)`

C

`(640+t)/(720)`

D

`(640-t)/(720)`

Text Solution

Verified by Experts

The correct Answer is:
D
`m xx 80+ mxx1 xx t=(M-m) xx1xx(100-t)+540(M-m)`
`720 m=(640-t)M`.
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