# Cyclic Processes and the Carnot Cycle

Physics II · Thermodynamics · https://tryals.app/learn/physics-ii/cyclic-processes-and-the-carnot-cycle

## Going Round in Circles Usefully

A **cyclic** process returns the system to its starting state, so $\Delta U = 0$ over the cycle and

$$W_{net} = Q_{in} - Q_{out}$$

An engine cannot simply convert all its heat into work, it must **dump** some to a cold reservoir. Efficiency is what you get over what you pay:

$$\eta = \frac{W_{net}}{Q_{in}} = 1 - \frac{Q_{out}}{Q_{in}}$$

The **Carnot cycle** consists of four reversible steps: isothermal expansion at $T_h$ absorbing $Q_h$, adiabatic expansion cooling to $T_c$, isothermal compression at $T_c$ rejecting $Q_c$, and adiabatic compression back. Because every step is reversible, the heats are in the ratio of the temperatures, giving

$$\eta_{Carnot} = 1 - \frac{T_c}{T_h}$$

**Carnot's theorem** states that no engine working between two reservoirs can beat this, and that all reversible engines between the same pair achieve exactly it, regardless of working substance. That substance-independence is what lets the ratio $Q_c/Q_h$ define the **thermodynamic temperature scale**, with no reference to any material at all.

The bound is severe. A power station with $T_h = 800$ K and $T_c = 300$ K is capped at $1 - 300/800 = 62.5\%$, and real plants reach perhaps 40%. Raising efficiency means raising $T_h$ or lowering $T_c$, and the cold reservoir is usually the environment.

Run in reverse, the same cycle becomes a **refrigerator** or **heat pump**, characterised by a coefficient of performance rather than an efficiency: $\text{COP}_{fridge} = T_c/(T_h - T_c)$, which can exceed 1, moving heat is cheaper than creating it.

> **Common pitfall:** using Celsius in the Carnot formula. The ratio $T_c/T_h$ is meaningful only on an absolute scale; using Celsius for a cycle between 27 °C and 227 °C gives $1 - 27/227 = 88\%$ instead of the correct $1 - 300/500 = 40\%$.

## Practice questions

8 of this lesson's 11 practice questions, with answers. The full set is in the app.

### 1. Complete the account of why a heat engine cannot be perfectly efficient.

**Answer:** A cyclic engine must **reject** some heat to a cold reservoir in order to close its cycle, so its efficiency is capped even when it is perfectly **frictionless**. This limit is imposed by the **second** law, and the maximum is set by the ratio of the two **absolute** temperatures.

**Why:** The cap is imposed by the second law, not by friction: even a perfect Carnot engine must reject heat to return to its starting state. Friction makes real engines worse still, but removing it entirely would not lift the ceiling.

Page: https://tryals.app/practice/physics-ii/cyclic-processes-and-the-carnot-cycle/complete-the-account-of-why-a-heat-engine-cannot-be-perfectly

### 2. An engine absorbs 900 J from a hot reservoir and rejects 600 J to a cold one per cycle. Compute its efficiency as a percentage, to the nearest whole number.

**Answer:** 33 (within ±1)

**Why:** $\eta = (Q_h - Q_c)/Q_h = 300/900 = 0.333$, so **33 %**. A third of the absorbed heat becomes work and two thirds is dumped.

Page: https://tryals.app/practice/physics-ii/cyclic-processes-and-the-carnot-cycle/an-engine-absorbs-900-j-from-a-hot-reservoir-and-rejects-600-j-to-a

### 3. All reversible engines operating between the same two reservoirs have the same efficiency, whatever their working substance.

**Answer:** True

**Why:** True. Carnot’s theorem makes the efficiency depend only on the two temperatures. That substance-independence is precisely what lets $Q_c/Q_h$ define an absolute temperature scale with no material reference.

Page: https://tryals.app/practice/physics-ii/cyclic-processes-and-the-carnot-cycle/all-reversible-engines-operating-between-the-same-two-reservoirs-have

### 4. Arrange the four steps of the Carnot cycle in order, beginning at the hot reservoir.

**Answer:**

1. Isothermal expansion at the hot temperature, absorbing heat
2. Adiabatic expansion, cooling to the cold temperature
3. Isothermal compression at the cold temperature, rejecting heat
4. Adiabatic compression, warming back to the hot temperature

**Why:** The cycle alternates: absorb heat isothermally hot, coast down adiabatically, reject heat isothermally cold, and climb back adiabatically. Only the isothermal legs exchange heat, and the adiabatic legs merely change temperature.

Page: https://tryals.app/practice/physics-ii/cyclic-processes-and-the-carnot-cycle/arrange-the-four-steps-of-the-carnot-cycle-in-order-beginning-at-the

### 5. Carnot efficiency depends strictly on reservoir temperatures rather than the working substance. Why does this independence allow the cycle to establish the absolute thermodynamic temperature scale?

A. It ensures heat exchanges scale linearly across any gas law.
B. It offers a scale independent of any material properties.
C. It guarantees that internal energy remains fixed at all steps.
D. It eliminates heat rejection across real engineering cycles.

**Answer:** B. It offers a scale independent of any material properties.

**Why:** Thermometers reliant on substance properties fail when materials change phase or depart from ideal behaviour. Reversible heat ratios bypass material quirks entirely, grounding absolute zero dynamically.

Page: https://tryals.app/practice/physics-ii/cyclic-processes-and-the-carnot-cycle/carnot-efficiency-depends-strictly-on-reservoir-temperatures-rather

### 6. Which statements about cyclic processes are correct?

A. The net work equals the difference between heat absorbed and heat rejected
B. A cycle can convert all absorbed heat into work if it is frictionless
C. The internal energy change over a complete cycle is zero
D. The net work is the area enclosed by the cycle on a P-V diagram

**Answer:** A. The net work equals the difference between heat absorbed and heat rejected; C. The internal energy change over a complete cycle is zero; D. The net work is the area enclosed by the cycle on a P-V diagram

**Why:** Returning to the same state makes $\Delta U = 0$, so the net work is the difference of the heats, and it appears as the enclosed area. Full conversion is forbidden by the second law regardless of friction.

Page: https://tryals.app/practice/physics-ii/cyclic-processes-and-the-carnot-cycle/which-statements-about-cyclic-processes-are-correct

### 7. A Carnot engine has a hot reservoir at 600 K and an efficiency of 25 %. Compute the cold reservoir temperature in K.

**Answer:** 450 (within ±5)

**Why:** $\eta = 1 - T_c/T_h$ gives $T_c = T_h(1 - \eta) = 600 \times 0.75 = 450$ K. Note how modest an efficiency a 150 K temperature drop buys.

Page: https://tryals.app/practice/physics-ii/cyclic-processes-and-the-carnot-cycle/a-carnot-engine-has-a-hot-reservoir-at-600-k-and-an-efficiency-of-25

### 8. Match each device to what it does with heat and work.

**Answer:**

- Heat engine → Takes heat from hot, delivers work, dumps heat to cold
- Refrigerator → Uses work to move heat from cold to hot
- Heat pump → Uses work to deliver heat into the warm space
- Perpetual motion of the second kind → Would convert heat wholly to work, and is impossible

**Why:** The engine and the reversed cycle are the same four steps; what differs is which flow you count as the useful output. The forbidden device is the one that takes heat from a single reservoir and turns it entirely into work.

Page: https://tryals.app/practice/physics-ii/cyclic-processes-and-the-carnot-cycle/match-each-device-to-what-it-does-with-heat-and-work