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1 gram of ice is mixed with 1 gram of st...

1 gram of ice is mixed with 1 gram of steam. At thermal equilibrium, the temperature of the mixture is

A

`55^(@)C`

B

`50^(@)C`

C

`100^(@)C`

D

`0^(@)C`

Text Solution

Verified by Experts

The correct Answer is:
C
Energy supplied by 1gm of steam `=420J + 2260J = 2345J`
Energy required to convert 1gm of ice into water at `O^(@)C = 335J`
Remaining energy (2345-335)J is used to raise the temperature of water beyond `100^(@)C`
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