Question

the heat capacity of 0.125kg of water is measured to 523jk^-1 at room temperature, hence calculate the heat capacity of water. (a) per unit mass (b) per unit volume​

Answer 1

Explanation:

The heat capacity of 0.125 kg of water is measured to be 523 JK at room temperature Hence calculate the heat capacity of water (a) per unit mass and (b) per unit volume, 3. Define the following terms as applicable to thermodynamics system, thermal equilibrium, thermally isolated system, extensive function of state and intensive function of state

Answer 2

Given: The heat capacity of 0.125 kg of water is measured to 523 JK^-1 at room temperature.

To find: The heat capacity of water (a) per unit mass (b) per unit volume​

Solution:

• Now we have given that the heat capacity of 0.125 kg of water is measured to 523 jk^-1 at room temperature.

• So heat capacity per unit mass c is given by :

                 heat capacity/mass

• So putting the values in formula, we get:

                 c = 523 JK^-1 / 0.125 kg

                 c = 4.184 x 10^3 JK^-1 kg^-1

• Now  heat capacity per unit volume C is given by :

                 c x density of water

• So putting the values in formula, we get:

                 C = 4.184 x 10^3 JK^-1 kg^-1 x 1000 kg m^-3

                 C = 4.184 x 10^6 J K^-1 m^-3

Answer:

            So the heat capacity of water

            (a) per unit mass is 4.184 x 10^3 JK^-1 kg^-1 and  

            (b) per unit volume​ is 4.184 x 10^6 J K^-1 m^-3