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2026 National Preliminary physics Topic 27 Free

Heat engines, efficiency and thermodynamic cycles

Thermal efficiency of a carnot engine: eta = 1 - tc/th · Sub-topic 1

PRELIMINARY STAGE 2026

Konongo Odumasi Senior High School: 45 points (Winner, qualified to one-eighth stage)

Our Lady of Mount Carmel Girls' Senior High School: 31 points

Jaben Senior High School: 28 points


QUESTION

Find the thermal efficiency of a Carnot engine that operates between the given pair of bath temperatures.

1. 25°C and 95°C.

ANSWER: 19.0%

SOLUTION 1:

$T_c = 25 + 273.15$

$T_c = 298.15 \text{ K}$

$T_h = 95 + 273.15$

$T_h = 368.15 \text{ K}$

$\eta = 1 - \dfrac{T_c}{T_h}$

$\eta = 1 - \dfrac{298.15}{368.15}$

$\eta \approx 0.1901$

$\eta \approx 19.0\%$

SOLUTION 2:

$T_c = 25 + 273$

$T_c = 298 \text{ K}$

$T_h = 95 + 273$

$T_h = 368 \text{ K}$

$\eta = 1 - \dfrac{T_c}{T_h}$

$\eta = 1 - \dfrac{298}{368}$

$\eta \approx 0.1902$

$\eta \approx 19.0\%$

2. 15°C and 125°C.

ANSWER: 27.6%

SOLUTION 1:

$T_c = 15 + 273.15$

$T_c = 288.15 \text{ K}$

$T_h = 125 + 273.15$

$T_h = 398.15 \text{ K}$

$\eta = 1 - \dfrac{T_c}{T_h}$

$\eta = 1 - \dfrac{288.15}{398.15}$

$\eta \approx 0.2763$

$\eta \approx 27.6\%$

SOLUTION 2:

$T_c = 15 + 273$

$T_c = 288 \text{ K}$

$T_h = 125 + 273$

$T_h = 398 \text{ K}$

$\eta = 1 - \dfrac{T_c}{T_h}$

$\eta = 1 - \dfrac{288}{398}$

$\eta \approx 0.2764$

$\eta \approx 27.6\%$

3. 27°C and 82°C.

ANSWER: 15.5%

SOLUTION 1:

$T_c = 27 + 273.15$

$T_c = 300.15 \text{ K}$

$T_h = 82 + 273.15$

$T_h = 355.15 \text{ K}$

$\eta = 1 - \dfrac{T_c}{T_h}$

$\eta = 1 - \dfrac{300.15}{355.15}$

$\eta \approx 0.1549$

$\eta \approx 15.5\%$

SOLUTION 2:

$T_c = 27 + 273$

$T_c = 300 \text{ K}$

$T_h = 82 + 273$

$T_h = 355 \text{ K}$

$\eta = 1 - \dfrac{T_c}{T_h}$

$\eta = 1 - \dfrac{300}{355}$

$\eta \approx 0.1549$

$\eta \approx 15.5\%$


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PRACTICE QUESTIONS

1. QUESTION: Find the thermal efficiency of a Carnot engine that operates between baths at $3.00\times10^{2}$ K and $4.00\times10^{2}$ K.

ANSWER: 25.0%

SOLUTION:

$\eta = 1 - \dfrac{T_c}{T_h}$

$\dfrac{T_c}{T_h} = \dfrac{3.00\times10^{2}}{4.00\times10^{2}}$

$\dfrac{T_c}{T_h} = \dfrac{3\times10^{2}}{4\times10^{2}}$

$\dfrac{T_c}{T_h} = \dfrac{3}{4}$

$\dfrac{T_c}{T_h} = 0.750$

$\eta = 1 - 0.750$

$\eta = 0.250$

$\eta = 25.0\%$


2. QUESTION: Find the thermal efficiency of a Carnot engine that operates between baths at $2.00\times10^{2}$ K and $5.00\times10^{2}$ K.

ANSWER: 60.0%

SOLUTION:

$\eta = 1 - \dfrac{T_c}{T_h}$

$\dfrac{T_c}{T_h} = \dfrac{2.00\times10^{2}}{5.00\times10^{2}}$

$\dfrac{T_c}{T_h} = \dfrac{2\times10^{2}}{5\times10^{2}}$

$\dfrac{T_c}{T_h} = \dfrac{2}{5}$

$\dfrac{T_c}{T_h} = 0.400$

$\eta = 1 - 0.400$

$\eta = 0.600$

$\eta = 60.0\%$


3. QUESTION: Find the thermal efficiency of a Carnot engine that operates between baths at $3.00\times10^{2}$ K and $6.00\times10^{2}$ K.

ANSWER: 50.0%

SOLUTION:

$\eta = 1 - \dfrac{T_c}{T_h}$

$\dfrac{T_c}{T_h} = \dfrac{3.00\times10^{2}}{6.00\times10^{2}}$

$\dfrac{T_c}{T_h} = \dfrac{3\times10^{2}}{6\times10^{2}}$

$\dfrac{T_c}{T_h} = \dfrac{1}{2}$

$\dfrac{T_c}{T_h} = 0.500$

$\eta = 1 - 0.500$

$\eta = 0.500$

$\eta = 50.0\%$


4. QUESTION: Find the thermal efficiency of a Carnot engine that operates between baths at $3.00\times10^{2}$ K and $8.00\times10^{2}$ K.

ANSWER: 62.5%

SOLUTION:

$\eta = 1 - \dfrac{T_c}{T_h}$

$\dfrac{T_c}{T_h} = \dfrac{3.00\times10^{2}}{8.00\times10^{2}}$

$\dfrac{T_c}{T_h} = \dfrac{3\times10^{2}}{8\times10^{2}}$

$\dfrac{T_c}{T_h} = \dfrac{3}{8}$

$\dfrac{T_c}{T_h} = 0.375$

$\eta = 1 - 0.375$

$\eta = 0.625$

$\eta = 62.5\%$


5. QUESTION: Find the thermal efficiency of a Carnot engine that operates between baths at $4.00\times10^{2}$ K and $5.00\times10^{2}$ K.

ANSWER: 20.0%

SOLUTION:

$\eta = 1 - \dfrac{T_c}{T_h}$

$\dfrac{T_c}{T_h} = \dfrac{4.00\times10^{2}}{5.00\times10^{2}}$

$\dfrac{T_c}{T_h} = \dfrac{4\times10^{2}}{5\times10^{2}}$

$\dfrac{T_c}{T_h} = \dfrac{4}{5}$

$\dfrac{T_c}{T_h} = 0.800$

$\eta = 1 - 0.800$

$\eta = 0.200$

$\eta = 20.0\%$


6. QUESTION: Find the thermal efficiency of a Carnot engine that operates between baths at $5.00\times10^{2}$ K and $8.00\times10^{2}$ K.

ANSWER: 37.5%

SOLUTION:

$\eta = 1 - \dfrac{T_c}{T_h}$

$\dfrac{T_c}{T_h} = \dfrac{5.00\times10^{2}}{8.00\times10^{2}}$

$\dfrac{T_c}{T_h} = \dfrac{5\times10^{2}}{8\times10^{2}}$

$\dfrac{T_c}{T_h} = \dfrac{5}{8}$

$\dfrac{T_c}{T_h} = 0.625$

$\eta = 1 - 0.625$

$\eta = 0.375$

$\eta = 37.5\%$


7. QUESTION: Find the thermal efficiency of a Carnot engine that operates between baths at $3.00\times10^{2}$ K and $1.00\times10^{3}$ K.

ANSWER: 70.0%

SOLUTION:

$\eta = 1 - \dfrac{T_c}{T_h}$

$\dfrac{T_c}{T_h} = \dfrac{3.00\times10^{2}}{1.00\times10^{3}}$

$\dfrac{T_c}{T_h} = \dfrac{3\times10^{2}}{10^{3}}$

$\dfrac{T_c}{T_h} = 0.300$

$\eta = 1 - 0.300$

$\eta = 0.700$

$\eta = 70.0\%$


8. QUESTION: Find the thermal efficiency of a Carnot engine that operates between baths at $2.50\times10^{2}$ K and $4.00\times10^{2}$ K.

ANSWER: 37.5%

SOLUTION:

$\eta = 1 - \dfrac{T_c}{T_h}$

$\dfrac{T_c}{T_h} = \dfrac{2.50\times10^{2}}{4.00\times10^{2}}$

$\dfrac{T_c}{T_h} = \dfrac{25\times10^{1}}{4\times10^{2}}$

$\dfrac{T_c}{T_h} = \dfrac{25}{4} \times10^{-1}$

$\dfrac{T_c}{T_h} = 0.625$

$\eta = 1 - 0.625$

$\eta = 0.375$

$\eta = 37.5\%$


9. QUESTION: Find the thermal efficiency of a Carnot engine that operates between baths at $4.00\times10^{2}$ K and $1.60\times10^{3}$ K.

ANSWER: 75.0%

SOLUTION:

$\eta = 1 - \dfrac{T_c}{T_h}$

$\dfrac{T_c}{T_h} = \dfrac{4.00\times10^{2}}{1.60\times10^{3}}$

$\dfrac{T_c}{T_h} = \dfrac{4\times10^{2}}{16\times10^{2}}$

$\dfrac{T_c}{T_h} = \dfrac{1}{4}$

$\dfrac{T_c}{T_h} = 0.250$

$\eta = 1 - 0.250$

$\eta = 0.750$

$\eta = 75.0\%$