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Question

10 gm of ice cubes at 0C are released in a tumbler (water equivalent 55 gm) at 40C. Assuming that negligible heat is taken from the surroundings, the temperature of water in the tumbler becomes nearly (L=80 cal/g)
  1. 31.54C
  2. 21.54C
  3. 19C
  4. 15.68C


Solution

The correct option is B 21.54C
Let the final temperature be T
Heat gained by ice is heat required to melt ice to water at 0C and heat required to increase temperature of water from 0C to TC
=mL+m×s×(T0)
=10×80+10×1×T
Heat lost by tumbler =55×1×(40T)
By using law of calorimetry,
800+10T=55×(40T)
T=21.54C

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