Path-independence of state properties: A system goes from state 1 to state 3 via path 1–2–3 with Q = +100 kJ and W = +60 kJ. Along an alternative path 1–4–3, the work is W = +20 kJ. What heat Q is absorbed along 1–4–3 (kJ)?

Introduction / Context:This problem tests recognition that internal energy is a state property and therefore path independent. While heat and work depend on the path, their difference (for a closed system neglecting KE/PE changes) equals the change in internal energy between the same end states.

Given Data / Assumptions:

• Closed system; sign convention: heat to system positive, work by system positive.
• Path 1–2–3: Q_123 = +100 kJ, W_123 = +60 kJ.
• Path 1–4–3: W_143 = +20 kJ; Q_143 = ?

Concept / Approach:First law for a closed system: ΔU = Q − W. Since states 1 and 3 are the same regardless of path, ΔU_123 = ΔU_143. Compute ΔU from the known path, then solve for the unknown heat on the alternative path using the same ΔU value.

Step-by-Step Solution:

Verification / Alternative check:Check consistency: Larger work on the first path (60 kJ) required larger heat input (100 kJ) to reach the same ΔU; with smaller work (20 kJ) on the second path, proportionally smaller heat (60 kJ) is needed to maintain ΔU = +40 kJ.

Why Other Options Are Wrong:

• −140, −80, −40 kJ: Incorrect sign/magnitude; they would imply a different ΔU, violating state-function path independence.

Common Pitfalls:Using Q + W instead of Q − W; forgetting the sign convention (work by system positive).

Final Answer:+60 kJ