1.

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

Answer»

`55^(@)C`
`50^(@)C`
`100^(@)C`
`0^(@)C`

Solution :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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