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Steam at 100^(@)C is passed into 20 g of...

Steam at `100^(@)C` is passed into `20 g` of water at `10^(@)C` when water acquire a temperature of `80^(@)C`, the mass of water present will be
[Take specific heat of water `= 1 cal g^(-1).^(@) C^(-1)` and latent heat of steam ` = 540 cal g^(-1)`]

A

`24 g`

B

`31.5 g`

C

`42.5 g`

D

`22.5 g`

Text Solution

Verified by Experts

The correct Answer is:
D
Heat gained by water= heat lost by steam
`20xx1(80-10)`
`= mxx540+mxx1xx(100-80)`
or `1400= 560m or m= 2.5 g`
Total mass of water `= 20+2.5= 22.5g`
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