MediumPhysicsHigh School (11–12)Heat & Thermodynamics › Heat Engines, the Second Law and EntropyNumerical

A gas goes round a cycle of three steps.

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A gas goes round a cycle of three steps. A to B: it takes in $520\ \text{J}$ of heat at a steady volume. B to C: it takes in $380\ \text{J}$ of heat and does $380\ \text{J}$ of work. C to A: it gives out $760\ \text{J}$ of heat while $240\ \text{J}$ of work is done on it. Find $\Delta U$ for each step, the net work, the net heat and the efficiency.

Need a reminder? The key ideaShow

In a cycle the gas returns to its start, so $\Delta U = 0$ and the net heat equals the net work. The net work is the area of the loop on a pressure-volume graph: positive for a clockwise loop, negative for an anticlockwise one.

The first law: $\Delta U = Q - W$ . $Q$ is positive when heat goes into the gas; $W$ is positive when the gas does work on its surroundings, so work done on the gas is negative. For an ideal gas, $\Delta U > 0$ means it gets warmer.

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

← All Problems

← Say whether each process is nearly reversible or irreversible, and give a reason.An engine takes 4200 J of heat from its hot reservoir each cycle and gives 2940 J to the cold one. →

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