# Heat engine (application of the second law of thermodynamics) – problems and solutions

30 Heat engine (application of the second law of thermodynamics) – problems and solutions

1. Heat input QH = 3000 Joule and heat output QL = 1000 Joule. What is the efficiency of the heat engine?

Known :

Heat input (QH) = 3000 Joule

Heat output (QL) = 1000 Joule

Work done by the engine (W) = 3000 – 1000 = 2000 Joule

Wanted: Efficiency of the heat engine (e)

Solution :

2. A heat engine produces 2000 Joule of mechanical work and the engine discharges heat to the environment at a rate of 500 Joule. What is the efficiency of the heat engine?

Known :

The work is done by the engine (W) = 2000 Joule

Heat output (QL) = 5000 Joule

Heat input (QH) = 2000 + 5000 = 7000 Joule

Wanted: Efficiency (e)

Solution :

3. A heat engine has an efficiency of 30%. If the engine produces 10,000 Joule of mechanical work, how much heat is discharged as waste heat from this engine?

Known :

Efficiency (e) = 30 % = 30/100 = 0.3

The work is done by engine (W) = 3,000 Joule

Wanted : Heat output (QL)

Solution :

Heat output (QL) = heat input (QH) – work is done by engine (W)

QL = 10,000 Joule – 3,000 Joule

QL = 7,000 Joule

4. Determine the efficiency of a Carnot engine operating between two reservoirs at temperatures of 500 K and 300 K.
Solution: Using the Carnot efficiency formula, $$\eta = 1 – \frac{T_{\text{cold}}}{T_{\text{hot}}}$$, we get $$\eta = 1 – \frac{300}{500} = 0.4$$ or $$40\%$$.

See also  Intensity of sound – problems and solutions

5. A heat engine absorbs 2000 J of heat from a hot reservoir and expels 1200 J to a cold reservoir. Calculate the work done by the engine.
Solution: Using the first law of thermodynamics, $$Q_{\text{hot}} = W + Q_{\text{cold}}$$, the work done is $$W = 2000 – 1200 = 800\ \text{J}$$.

6. Calculate the efficiency of a heat engine that absorbs 1000 J from a hot reservoir and does 600 J of work.
Solution: $$\eta = \frac{W}{Q_{\text{hot}}} = \frac{600}{1000} = 0.6$$ or $$60\%$$.

7. A heat engine operates between 400 K and 200 K. Determine the maximum possible efficiency.
Solution: $$\eta_{\text{max}} = 1 – \frac{T_{\text{cold}}}{T_{\text{hot}}} = 1 – \frac{200}{400} = 0.5$$ or $$50\%$$.

8. A heat engine rejects 300 J to the cold reservoir while doing 200 J of work. Calculate the heat absorbed from the hot reservoir.
Solution: $$Q_{\text{hot}} = W + Q_{\text{cold}} = 200 + 300 = 500\ \text{J}$$.

9. Determine the efficiency of an Otto cycle with a compression ratio of 8.
Solution: $$\eta = 1 – \frac{1}{r^{(\gamma – 1)}} = 1 – \frac{1}{8^{(1.4 – 1)}} \approx 0.564$$ or $$56.4\%$$, where $$\gamma$$ is the heat capacity ratio.

10. A heat engine performs 150 J of work and rejects 100 J to the cold reservoir. Find the heat absorbed from the hot reservoir.
Solution: $$Q_{\text{hot}} = W + Q_{\text{cold}} = 150 + 100 = 250\ \text{J}$$.

11. Determine the efficiency of a Diesel engine with a compression ratio of 18 and a cutoff ratio of 2.
Solution: $$\eta = 1 – \frac{1}{r^{(\gamma – 1)}} \left( \frac{r_c – 1}{r_c} \right) \approx 0.627$$ or $$62.7\%$$, where $$r$$ is the compression ratio, $$r_c$$ is the cutoff ratio, and $$\gamma$$ is the heat capacity ratio.

See also  Uniform circular motion - problems and solutions

12. Calculate the work done by a heat engine that absorbs 400 J and has an efficiency of 50%.
Solution: $$W = \eta \cdot Q_{\text{hot}} = 0.5 \cdot 400 = 200\ \text{J}$$.

13. A heat engine operates between 600 K and 300 K. Find the Carnot efficiency.
Solution: $$\eta = 1 – \frac{T_{\text{cold}}}{T_{\text{hot}}} = 1 – \frac{300}{600} = 0.5$$ or $$50\%$$.

14. A heat engine absorbs 3000 J and rejects 1800 J to the cold reservoir. Calculate the work done.
Solution: $$W = Q_{\text{hot}} – Q_{\text{cold}} = 3000 – 1800 = 1200\ \text{J}$$.

15. Determine the efficiency of a Stirling engine operating between two reservoirs at 800 K and 400 K.
Solution: $$\eta = 1 – \frac{T_{\text{cold}}}{T_{\text{hot}}} = 1 – \frac{400}{800} = 0.5$$ or $$50\%$$.

16. A heat engine rejects 500 J to the cold reservoir while doing 300 J of work. Calculate the heat absorbed from the hot reservoir.
Solution: $$Q_{\text{hot}} = W + Q_{\text{cold}} = 300 + 500 = 800\ \text{J}$$.

17. Determine the efficiency of an Ericsson cycle with hot and cold reservoirs at 1000 K and 250 K.
Solution: $$\eta = 1 – \frac{T_{\text{cold}}}{T_{\text{hot}}} = 1 – \frac{250}{1000} = 0.75$$ or $$75\%$$.

18. A heat engine does 700 J of work and has an efficiency of 70%. Calculate the heat rejected to the cold reservoir.
Solution: $$Q_{\text{cold}} = \frac{W}{\eta} – W = \frac{700}{0.7} – 700 \approx 1000 – 700 = 300\ \text{J}$$.

See also  Centripetal force in uniform circular motion – problems and solutions

19. Determine the efficiency of a Brayton cycle with pressure ratios of 6.5 and $$\gamma = 1.4$$.
Solution: $$\eta = 1 – \frac{1}{r_p^{(\gamma – 1)/\gamma}} \approx 0.433$$ or $$43.3\%$$, where $$r_p$$ is the pressure ratio.

20. Calculate the work done by a heat engine that absorbs 250 J and rejects 100 J to the cold reservoir.
Solution: $$W = Q_{\text{hot}} – Q_{\text{cold}} = 250 – 100 = 150\ \text{J}$$.

21. A heat engine operates between 450 K and 150 K. Find the Carnot efficiency.
Solution: $$\eta = 1 – \frac{T_{\text{cold}}}{T_{\text{hot}}} = 1 – \frac{150}{450} \approx 0.667$$ or $$66.7\%$$.

22. Determine the efficiency of an Atkinson cycle with a compression ratio of 10.
Solution: $$\eta = 1 – \frac{1}{r^{(\gamma – 1)}} \approx 0.593$$ or $$59.3\%$$,

where $$r$$ is the compression ratio and $$\gamma$$ is the heat capacity ratio.

20. A heat engine absorbs 180 J and has an efficiency of 40%. Calculate the work done.
Solution: $$W = \eta \cdot Q_{\text{hot}} = 0.4 \cdot 180 = 72\ \text{J}$$.

Note: For problems involving specific heat cycles (like Otto, Diesel, etc.), certain values like heat capacity ratio $$\gamma$$ were used, which are typical for certain gases (e.g., $$\gamma = 1.4$$ for air). These values might differ depending on the specific substances used in the engine.