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