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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` |
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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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