The Carnot Cycle (DP IB Physics: HL): Revision Note

In an idealised heat engine, the hot and cold reservoirs are held at temperatures of $T_H$ and $T_C$ respectively.

Using the equation for the change in entropy

$∆S=\frac{∆Q}{T}$

Show that the maximum theoretical efficiency of a heat engine is given by

${\eta }_C=1-\frac{T_C}{T_H}$

Answer:

Step 1: Determine the change in entropy during isothermal expansion

• In isothermal expansion (AB):  the gas absorbs heat $Q_H$ from a hot reservoir at temperature $T_H$

• Therefore, the increase in entropy is:

$∆S_{AB}=\frac{Q_H}{T_H}$

Step 2: Determine the change in entropy during isothermal compression

• In isothermal compression (CD):  the gas transfers heat $Q_C$ to a cold reservoir at temperature $T_C$

• Therefore, the decrease in entropy is:

$∆S_{CD}=-\frac{Q_C}{T_C}$

Step 3: Consider the net change in entropy over the cycle

• During adiabatic expansion (BC) and compression (DA), entropy does not change as there is no thermal energy gained or lost

$∆S_{BC}=∆S_{DA}=0$

• We know that the overall entropy of the system does not change in a cyclic process, therefore

$∆S=∆S_{AB}+∆S_{BC}+∆S_{CD}+∆S_{DA}$

$∆S=\frac{Q_H}{T_H}+0+\left(-\frac{Q_C}{T_C}\right)+0=0$

$\frac{Q_H}{T_H}=\frac{Q_C}{T_C}\Rightarrow \frac{Q_C}{Q_H}=\frac{T_C}{T_H}$

Step 4: Substitute the expression into the equation for the efficiency of a heat engine

• The efficiency of a heat engine is given by:

$\eta =\frac{usefulwork}{inputenergy}=\frac{W}{Q_H}$

• Where useful work is $W=Q_H-Q_C$

$\eta =\frac{Q_H-Q_C}{Q_H}=1-\frac{Q_C}{Q_H}$

• Combining with the expression derived above gives:

${\eta }_C=1-\frac{T_C}{T_H}$

An engineer designs a heat engine that has an inlet temperature of 500 K and an outlet temperature of 300 K. The engineer claims that 100 kJ of thermal energy flows out of the hot reservoir and 25 kJ of thermal energy flows into the cold reservoir.

Determine, with reference to the second law of thermodynamics, whether this engine is thermodynamically possible.

Answer:

Step 1: Determine the efficiency of the proposed engine

• The efficiency of this engine would be

$\eta =\frac{W}{Q_H}=\frac{Q_H-Q_C}{Q_H}=1-\frac{Q_C}{Q_H}$

• Where:

  • Heat transferred in, $Q_H$ = 100 kJ

  • Heat transferred out, $Q_C$ = 25 kJ

Efficiency = $1-\frac{25}{100}$ = 0.75 = 75%

Step 2: Determine the maximum theoretical (Carnot) efficiency of the proposed engine

• A Carnot engine operating between the same temperatures would have an efficiency of

${\eta }_C=1-\frac{T_C}{T_H}$

• Where:

  • Inlet temperature, $T_H$ = 500 K

  • Outlet temperature, $T_C$ = 300 K

Carnot efficiency:  ${\eta }_C=1-\frac{300}{500}$ = 0.4 = 40%

Step 3: Discuss the proposed engine in relation to the second law

• The Clausius form of the second law states: it is impossible for heat to flow from a cooler body to a hotter body without performing work

• This law sets an upper limit on the maximum possible efficiency of the transfer of thermal energy to mechanical energy in a heat engine

• The maximum possible efficiency of the proposed engine is 40%, but the engineer is proposing an efficiency of 75% i.e. an efficiency greater than the Carnot efficiency

• This violates the Clausius form of the second law, hence the proposed engine is impossible