Discussion
If Q1 and Q2 are respectively, the heat supplied and expelled by a system at a constant temperature, then the work done by the system is
Answer: A
The work done is \(Q_{1} - Q_{2}\).
By the first law, \(Q = \Delta U + W\). At constant temperature (for an ideal gas) or over a complete cycle, \(\Delta U = 0\), so the work done equals the net heat taken in: heat supplied \(Q_{1}\) minus heat given out \(Q_{2}\).
\(Q_{1} + Q_{2}\) adds the heat expelled instead of subtracting it, and the halved forms have no basis in the first law.
This is the same balance used for a heat engine over a cycle, where the efficiency is \(\frac{Q_{1} - Q_{2}}{Q_{1}}\).
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