Sol.

Use the first law in the form Q – W = Δ E. Applied to process 1 → 2, we have

a – 100 = 100 ∴ a = 200 kJ

Applied to process 3 → 1, there results

100 – d = – 200 ∴ d = 300 kJ

The net work is then Σ W = W 1–2 + W 2–3 + W 3–1 = 100 – 50 + 300 = 350 kJ. The first law for a cycle demands that

Σ Q = Σ W 200 + b + 100 = 350 ∴ b = 50 kJ

Finally, applying the first law to process 2 → 3 provides

50 – (– 50) = c ∴ c = 100 kJ

Note that, for a cycle, ΣΔ E = 0; this, in fact, could have been used to determine the value of c:

ΣΔ E = 100 + c – 200 = 0 ∴ c = 100 kJ.