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


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