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Question

One gram of ice is mixed with one gram of steam. At thermal equilibrium the temperature of mixture is
  1. 100C
  2. 0C
  3. 55C
  4. 80C


Solution

The correct option is A 100C
Heat required to melt 1g of ice at 0C to water at 0C=1×80 cal.
Heat required to raise temperature of 1 g of water from 0 C to 100 C=1×1×100=100cal
Total heat required for maximum temperature of 100C = 80 + 100 = 180 cal
As one gram of steam gives 540 cal of heat when it is converted to water at 100C,
therefore,  temperature of the mixture would be 100C.

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