Q- 1 What is the heat capacity ?
Answer :- Amount of heat which required to raise the temperature of one gram substance by one degree Centigrade is called heat capacity of that substance.
example:-
heat capacity of water is 4.18 j /gC
That means one gram of water require heat 4.18 j to raise the temperature of water by one degree centigrade.
Q-2 Determine specific heat in J/g times degree celcius . Mass of substance 29.2g, initial temp 34.4degrees celcius, final temp 68.1degrees celcius, heat applied 436.9 j, what is the formula of hear capacity ?
Answer
Formula for the heat Q = M*C*delta T
M = mass of the substance in gram
C = specific hear capacity of substance
delta T = final temperature - initial temperature
from the problem
delta T = 68.1 - 34.4
=33.7 C
Mass M = 29.2 gram
Heat Q = 436.9j
plug the value in above formula we get
Q = M*C*delta T
436.9 j = 29.2 gram * C* 33.7degree Celsius
436.9 = 984.04gram degree celcius * C
divide by 984.04gram degree Celsius we get
0.44 j / gram degree celcius = C (specific heat capacity )
Answer heat capacity = 0.44 j/g K or 0.44j/gC
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| heat capacity of water |
