Carnot cycle – problems and solutions

Carnot cycle – problems and solutions

1. If heat absorbed by the engine (Q₁) = 10,000 Joule, what is the work done by the Carnot engine?

Known:

Low temperature (T₂) = 400 K

High temperature (T₁) = 800 K

Heat input (Q₁) = 10,000 Joule

Wanted: Work done by Carnot engine (W)

Solution:

The efficiency of the Carnot engine :

Work was done by Carnot engine :

W = e Q₁

W = (1/2)(10,000) = 5000 Joule

2.

Based on graph above, what is the work done by engine in a cycle?

Known :

Low temperature (T_L) = 400 K

High temperature (T_H) = 600 K

Heat input (Q₁) = 600 Joule

Wanted: Work was done by Carnot engine (W)

Solution :

The efficiency of the Carnot engine :

Work done by Carnot engine :

W = e Q₁

W = (1/3)(600) = 200 Joule

3. Based on the graph below, what is the efficiency of the Carnot engine?

Known :

Low temperature (T_L) = 350 K

High temperature (T_H) = 500 K

Wanted : Efficiency of Carnot engine (e)

Solution :

Efficiency of Carnot engine :

e = (T_H – T_L) / T_H

e = (500 – 350) / 500

e = 150 / 500

e = 0.3

e = 30/100 = 30 %

4. Based on graph below, the heat engine’s high temperature is 600 K and low temperature is 400 K. If the work done by engine is W, what is the heat output.

Known :

Low temperature (T_L) = 400 K

High temperature (T_H) = 600 K

Wanted : heat output (Q₂)

Solution :

Efficiency of Carnot engine :

e = (T_H – T_L) / T_H

e = (600 – 400) / 600

e = 200 / 600

e = 1/3

Work done by Carnot engine :

W = e Q₁

W = work done by engine, e = efficiency, Q1 = heat input

W = (1/3)(Q₁)

3W = Q₁

Heat output :

Q₂ = Q₁ – W

Q₂ = 3W – W

Q₂ = 2W

5. Based on graph below, if the heat output is 3000 Joule, what is the heat input.

Known :

Low temperature (T_L) = 500 K

High temperature (T_H) = 800 K

Heat output (Q₂) = 3000 Joule

Wanted : Heat input (Q₁)

Solution :

Efficiency of Carnot engine :

e = (T_H – T_L) / T_H

e = (800 – 500) /8600

e = 300 / 800

e = 3/8

Work done by Carnot engine :

W = e Q₁

W = (3/8)(Q₁)

8W/3 = Q₁

Q₂ = Q₁ – W

Q₂ = 8W/3 – 3W/3

Q₂ = 5W/3

3Q₂ = 5W

W = 3Q₂/5 = 3(3000)/5 = 9000/5 = 1800

Heat absorbed by engine :

Q₁ = W + Q₂ = 1800 + 3000 = 4800 Joule

6. An Carnot engine absorbs heat at high temperature 800 Kelvin and efficiency of the Carnot engine is 50%. What is the high temperature to increase efficiency to 80% if the low temperature kept constant.

Known :

If high temperature (T_H) = 800 K , efficiency (e) = 50% = 0.5

Wanted : High temperature (T_H) if efficiency (e) = 80% = 0.8

Solution :

Low temperature = 400 Kelvin

What is the high temperature (T_H) if efficiency (e) = 80 % ?

High temperature = 2000 Kelvin

7. A Carnot engine works at high temperature 600 Kelvin with the efficiency of 40%. If the efficiency of the engine is 75% and the low temperature kept constant, what is the high temperature?

Known :

If high temperature (T_H) = 600 K , efficiency (e) = 40% = 0.4

Wanted : High temperature (T_H) if efficiency (e) = 75% = 0.75

Solution :

High temperature (T_H) if efficiency (e) = 75 % ?

High temperature = 1440 Kelvin

1. What is the Carnot cycle? Answer: The Carnot cycle is a theoretical thermodynamic cycle that represents the most efficient reversible heat engine cycle possible. It consists of two isothermal processes and two adiabatic processes.
2. Why is the Carnot cycle considered an ideal cycle? Answer: The Carnot cycle is considered ideal because it represents the upper limit of efficiency for any heat engine. No real engine can be more efficient than a Carnot engine operating between the same two temperature reservoirs.
3. What are the four processes in a Carnot cycle? Answer: The four processes in a Carnot cycle are:
   1. Isothermal expansion at the high temperature T_H​.
   2. Adiabatic expansion (where the system is thermally insulated and cools down).
   3. Isothermal compression at the low temperature T_C​.
   4. Adiabatic compression (where the system is thermally insulated and heats up).
4. Why is there no actual heat engine that operates on the Carnot cycle? Answer: Real engines have irreversible losses, such as friction, and cannot maintain perfect insulation during the adiabatic processes. Furthermore, it would be impractical to achieve the infinitely slow isothermal processes required by the Carnot cycle.
5. What is the efficiency of a Carnot engine? Answer: The efficiency $\eta$ of a Carnot engine operating between two temperature reservoirs T_H​ (hot) and T_C​ (cold) is given by:
   η=1−T_C/T_H​​​

   where temperatures are in Kelvin.

6. Why can’t a Carnot engine have 100% efficiency? Answer: A Carnot engine’s efficiency is dependent on the temperature difference between the hot and cold reservoirs. To achieve 100% efficiency, the cold reservoir’s temperature would need to be absolute zero (0 Kelvin), which is unattainable in practice.
7. What is the significance of reversibility in the Carnot cycle? Answer: Reversibility ensures that there are no entropy-generating processes, which means the cycle can operate at maximum efficiency. Any irreversible process would decrease the cycle’s efficiency.
8. How is the Carnot cycle related to the second law of thermodynamics? Answer: The Carnot cycle underpins the Second Law by establishing an upper limit on the efficiency of heat engines. The Second Law asserts that no engine can be more efficient than a Carnot engine operating between the same two temperatures.
9. Why is it impossible to have isothermal processes in real-world applications exactly as they appear in the Carnot cycle? Answer: An isothermal process, as depicted in the Carnot cycle, requires an infinite amount of time, which is impractical in real-world applications. This is because to maintain the isothermal condition, heat transfer should take place infinitesimally slowly.

10. How does the Carnot cycle help engineers and scientists? Answer: The Carnot cycle provides a theoretical benchmark for the maximum possible efficiency of heat engines. By comparing real engines to the Carnot cycle, engineers and scientists can identify areas for improvement and understand the fundamental limits of their designs.