Question

A closed system of constant volume experiences
a temperature rise of 25°C when a certain
process occurs. The heat transferred in the
process is 30 kJ. The specific heat at constant
volume for the pure substance comprising the
system is 1.2 kJ/kg°C, and the system contains
the
2.5 kg. of this substance. Determine:
(i) The change in internal energy
(ii) The work done.​

Answer 1

Answer:

Explanation:

Temperature rise(T2-T1)=25°C

the heat transferred Q=30kJ

specific heat at constant volume cv=1.2kJ/kg°C

mass of the substance m=2.5kg

ΔU=m∫T2 = cvdT

          ∫ T1

    =2.5∫T2 1.2dT= 3.0*(T2-T1)

             T1

   =  3.0*2.5 = 75kJ

thus change in internal energy is 75kJ

according to the first law of thermodynamics,

Q=ΔU + W

30=75+ W

W=75-30= -45 kJ

thus work done is -45kJ

Hope this helps