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1 g of ice at 0^@C is mixed with 1 g of ...

1 g of ice at `0^@C` is mixed with 1 g of steam at `100^@C`. After thermal equilibrium is achieved, the temperature of the mixture is

A

`0^@c`

B

`50^@c`

C

`80^@c`

D

`100^@c`

Text Solution

Verified by Experts

The correct Answer is:
D
Heat taken by ice to melt at `0^@C` water
`Q=mL=Ixx80=80cal//g^@C`
Heat given by steam
`Q=mL=mxxsxxtrianglet=540+1xx100^@C`
`Q=640^@cal//^@C`
Because heat of steam more than heat of ice. So the temperature of the mixture is remain `100^@C`
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