Question

A system undergoes a change of state during which 100 kJ of heat is transferred to it and it does 50 kJ of work. The system is brought back to its original state through a process during which 120 kJ of heat is transferred to it. The work done by the system in kJ is

1. 50
2. 70
3. 170
4. 200

Answer 1

Correct Answer - Option 3 : 170

Concept:

From the first law of thermodynamic for a cyclic process,

\(\oint \delta Q = \oint \delta W\)

Calculation:

Given:

Q₁ = 100 kJ, W₁ = 50 kJ,  Q₂ = 120 kJ

The given process is cyclic so from the first law of thermodynamic we can write

\(\oint \delta Q = \oint \delta W\)

Q₁ + Q₂ = W₁ + W₂

100 + 120 = 50 + W₂

W₂ = 220 – 50 = 170 kJ       

∴ the work done by the system is 170 kJ.