1.

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 isA. `100^@C`B. `55^@C`C. `75^@C`D. `0^@C`

Answer» Correct Answer - A
Heat taken by 1 g ice in transformation from ice at `0^@C` to water at` 100^@C` is
`Q = mL + msDeltatheta`
`= (1)(80) + (1)(1)(100)`
= 180 cal
Mass of steam condensed to give this much heat is
`m = Q/L = 180/540 = 1/3g`
This is less than 1g or total mass of steam.
Therefore, whole steam is not condensed and
mixture temperature is `100^@C`.


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