ICSE Class 10 Physics 2013 Solved Paper

Question : 56

Total: 70

A calorimeter of mass 50g and specific heat capacity 0.42Jg−1‌∘C−1 contains some mass of water at 20∘C. A metal piece of mass 20g at 100∘C is dropped into the calorimeter. After stirring, the final temperature of the mixture is found to be 22∘C. Find the mass of water used in the calorimeter.
[specific heat capacity of the metal piece = 0.3Jg−1‌∘C−1
specific heat capacity of water =4.2Jg−1 ‌∘C−1]

Solution:

Heat energy given by metal piece
‌=m⋅c⋅∆T1
‌=20×0⋅3×(100−22)
‌=468‌Joule
Heat energy gained by water
=mw×cw×∆T2
‌=mw×4.2×(22−20)
‌=mw×8.4‌ Joule ‌
Heat energy gained by calorimeter
‌=mc×cc×∆T2
‌=50×0.42×(22−20)
‌=42‌ Joule ‌
By principle of calorimetry
‌ Heat lost ‌=‌ Heat gained ‌
Heat energy given by metal = Heat energy gained by water + Heat energy gained by calorimeter
468‌=mw×8.4+42
mw‌=‌

468−42

8.4

‌=50.7‌gm.

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