Thermodynamics • Topic 3 of 3

Second Law, Heat Engines & Refrigerators

The first law says energy is conserved, but it allows processes that never actually happen — heat flowing from a cold body to a hot one, or a single reservoir's heat being turned fully into work. The second law of thermodynamics rules these out by fixing the natural direction of thermal processes. It has two equivalent classic statements:

Kelvin–Planck statement: it is impossible to build an engine that, working in a cycle, takes heat from a single reservoir and converts all of it into work. Some heat must always be rejected.
Clausius statement: heat cannot flow on its own from a colder body to a hotter body; external work must be done to make this happen.

A heat engine is a device that converts heat into work in a repeating cycle. It absorbs heat $Q_1$ from a hot source (at $T_1$), converts part of it into useful work $W$, and rejects the remainder $Q_2$ to a cold sink (at $T_2$). By energy conservation $W=Q_1-Q_2$, and the efficiency is the fraction of input heat turned into work:

$\eta=\frac{W}{Q_1}=\frac{Q_1-Q_2}{Q_1}=1-\frac{Q_2}{Q_1}$.
Since $Q_2>0$ always (Kelvin–Planck), $\eta<1$: no engine can be 100% efficient.

The Carnot engine is an ideal, reversible engine and is the most efficient engine possible between two given temperatures. Its cycle has four reversible steps: isothermal expansion (heat $Q_1$ in at $T_1$), adiabatic expansion (cooling to $T_2$), isothermal compression (heat $Q_2$ out at $T_2$) and adiabatic compression (back to start). For the Carnot cycle the heat ratio equals the temperature ratio, so:

$\eta_{Carnot}=1-\frac{T_2}{T_1}$, with temperatures in kelvin.
Efficiency rises as the source gets hotter or the sink gets colder; $\eta=1$ only if $T_2=0\ \text{K}$, which is unattainable.

A refrigerator (or heat pump) is a heat engine run in reverse. Using external work $W$, it extracts heat $Q_2$ from a cold region (inside the fridge, at $T_2$) and dumps $Q_1=Q_2+W$ into the warmer room (at $T_1$). Its performance is measured by the coefficient of performance (COP), the heat removed per unit work:

$\beta=\frac{Q_2}{W}=\frac{Q_2}{Q_1-Q_2}$.
For an ideal (Carnot) refrigerator $\beta=\frac{T_2}{T_1-T_2}$.
Unlike efficiency, the COP can be greater than 1 — that is why a refrigerator can move more heat than the work it consumes.

Solved Examples

Worked Example

A Carnot engine works between a source at 500 K and a sink at 300 K. Find its efficiency.

Solution

Step 1: Use $\eta=1-\frac{T_2}{T_1}$ with temperatures in kelvin.
Step 2: Substitute: $\eta=1-\frac{300}{500}=1-0.6$.
Step 3: Compute: $\eta=0.4=40\%$.

Answer: $\eta=40\%$.

Worked Example

A heat engine absorbs 1000 J from a source and rejects 700 J to the sink in each cycle. Find the work done and the efficiency.

Solution

Step 1: Work done $W=Q_1-Q_2=1000-700=300\ \text{J}$.
Step 2: Efficiency $\eta=1-\frac{Q_2}{Q_1}=1-\frac{700}{1000}$.
Step 3: Compute: $\eta=0.30=30\%$.

Answer: $W=300\ \text{J}$, $\eta=30\%$.

Worked Example

A Carnot engine has an efficiency of 50% when the sink is at 300 K. Find the temperature of the source.

Solution

Step 1: $\eta=1-\frac{T_2}{T_1}$, so $\frac{T_2}{T_1}=1-\eta=1-0.5=0.5$.
Step 2: Therefore $T_1=\frac{T_2}{0.5}=\frac{300}{0.5}$.
Step 3: Compute: $T_1=600\ \text{K}$.

Answer: $T_1=600\ \text{K}$.

Worked Example

A refrigerator extracts 240 J of heat from its interior while consuming 60 J of electrical work in each cycle. Find its coefficient of performance and the heat rejected to the room.

Solution

Step 1: COP $\beta=\frac{Q_2}{W}=\frac{240}{60}=4$.
Step 2: Heat rejected $Q_1=Q_2+W=240+60$.
Step 3: Compute: $Q_1=300\ \text{J}$.

Answer: $\beta=4$; heat rejected $Q_1=300\ \text{J}$.

Worked Example

An ideal refrigerator maintains its interior at 250 K while the room is at 300 K. Find its coefficient of performance.

Solution

Step 1: For an ideal (Carnot) refrigerator $\beta=\frac{T_2}{T_1-T_2}$.
Step 2: Substitute $T_2=250\ \text{K}$, $T_1=300\ \text{K}$: $\beta=\frac{250}{300-250}=\frac{250}{50}$.
Step 3: Compute: $\beta=5$.

Answer: $\beta=5$.

Worked Example

A Carnot engine takes 2000 J of heat from a source at 400 K and rejects heat to a sink at 300 K. Find the work done per cycle.

Solution

Step 1: Carnot efficiency $\eta=1-\frac{T_2}{T_1}=1-\frac{300}{400}=0.25$.
Step 2: Work done $W=\eta\,Q_1=0.25\times2000$.
Step 3: Compute: $W=500\ \text{J}$ (and rejected heat $Q_2=Q_1-W=1500\ \text{J}$).

Answer: $W=500\ \text{J}$.

Key Points

Second law fixes the direction of heat flow: Kelvin–Planck (no engine fully converts heat from one reservoir to work) and Clausius (heat does not flow cold→hot by itself).
Heat engine: absorbs $Q_1$, rejects $Q_2$, does work $W=Q_1-Q_2$; efficiency $\eta=1-\frac{Q_2}{Q_1}<1$.
Carnot engine is the most efficient reversible engine: $\eta=1-\frac{T_2}{T_1}$ (temperatures in kelvin).
Carnot efficiency rises with a hotter source or colder sink; $\eta=1$ only if $T_2=0\ \text{K}$, which is unattainable.
Refrigerator (engine in reverse): COP $\beta=\frac{Q_2}{W}=\frac{Q_2}{Q_1-Q_2}$; ideal $\beta=\frac{T_2}{T_1-T_2}$, which can exceed 1.

Quick Check — 4 MCQs

Tap an option to check your answer0 / 4

Q1.The Kelvin–Planck statement of the second law says it is impossible to:

Explanation: An engine working in a cycle must reject some heat; it cannot turn all the heat from one reservoir into work.

Q2.The efficiency of a Carnot engine working between $T_1$ and $T_2$ is:

Explanation: For a Carnot engine $\frac{Q_2}{Q_1}=\frac{T_2}{T_1}$, so $\eta=1-\frac{T_2}{T_1}$.

Q3.The efficiency of any heat engine is always:

Explanation: Because some heat $Q_2>0$ is always rejected, $\eta=1-\frac{Q_2}{Q_1}<1$.

Q4.The coefficient of performance of an ideal refrigerator is:

Explanation: For an ideal refrigerator $\beta=\frac{Q_2}{W}=\frac{T_2}{T_1-T_2}$, which can be greater than 1.

Practice more MCQs → 📝 Assignment · 30 marks