Friday, August 26, 2011

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 ?

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
heat capacity of water
heat capacity of water

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