ICSE Class 10 Physics 2014 Solved Paper

Question : 28

Total: 66

50g of metal piece at 27∘C requires 2400J of heat energy so as to attain a temperature of 327∘C. Calculate the specific heat capacity of the metal.

Solution:

Given m=50g;∆t=327∘C−27∘C=300∘C‌, ‌ Q=2400J.
‌‌ Now, ‌‌‌Q=mc∆t
‌2400=50×c×300
‌c=‌

2400

50×300

=0.16J∕g‌∘C.

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