Carnot Engine Important Questions

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Very Short Answers Questions [1 Mark Questions]

Ques. If the temperature of the source is increased, the efficiency of the Carnot engine

1. Remains constant
2. Decreases
3. Increases
4. First, it increases and then becomes constant

Ans. The correct answer is c. Increases

Explanation: The efficiency of an engine is given by

µ = 1 - T₂/T₁ 

Where T₁ is the temperature of the source

Hence when the temperature of the source increases, the efficiency of the Carnot engine increases.

Ques. Even the Carnot engine cannot give 100% efficiency because we cannot

1. Prevent radiation
2. Reach absolute zero temperature
3. Find ideal sources
4. Eliminate friction

Ans. The correct answer is b. Reach absolute zero temperature

Explanation: The efficiency of the Carnot engine is given by

µ = 1 - T₂/T₁ 

This shows that the efficiency becomes 100% only when the source temperature T₁ is infinite or the sink temperature is zero. Both conditions are impossible.

Ques. For which combination of working temperatures the efficiency of Carnot’s engine is highest

1. 80 K, 60 K
2. 100 K, 80 K
3. 60 K, 40 K
4. 40 K, 20 K

Ans. The correct answer is d. 40 K, 20 K

Explanation: The efficiency of the Carnot engine is given by

µ = 1 - (T₂/T₁)

This shows that efficiency will be maximum when the ratio (T₂/T₁) is minimum i.e.

µ = 1 - (20/40) = 0.5

Ques. An example of a reversed heat engine is

1. Refrigerator
2. Heat pump
3. All of the above
4. None of the above

Ans. The correct answer is c. All of the above

Explanation: Refrigerators and Heat pumps are the types of reverse heat engines.

Ques. The efficiency of a Carnot engine that operates between the temperature of 450 K and 300 K is

1. 33%
2. 44%
3. 55%
4. 66%

Ans. The correct answer is a. 33%

Explanation: The efficiency of the Carnot engine is given by

µ = 1 - (T₂/T₁)

Where T₁ is the source temperature and T₂ is the sink temperature. Therefore

µ = 1 - (300/450) = 0.33 or 33%

Ques. If the door of a refrigerator is kept open, then which of the following is true

1. Room is cooled
2. Room is heated
3. The room is either cooled or heated
4. The room is neither cooled nor heated

Ans. The correct answer is b. Room is heated

Explanation: In a refrigerator, the heat dissipated in the atmosphere is more than that taken from the cooling chamber, therefore the room is heated if the door of a refrigerator is kept open.

Ques. The Carnot engine is a theoretical thermodynamic cycle proposed by

1. Leonard Carnot
2. Kepler
3. Faraday
4. Newton

Ans. The correct answer is a. Leonard Carnot

Explanation: Leonard Carnot proposed the Carnot engine, a theoretical thermodynamic cycle. It estimates the maximum possible efficiency of a heat engine during the process of converting heat into work and, conversely, working between two reservoirs.

Ques. The Carnot’s engine’s efficiency depends on the absolute temperature of the

1. Sink
2. Source
3. Both source and sink
4. Neither source nor sink

Ans. The correct answer is c. Both source and sink

Explanation: The absolute temperature of the source and sink determines the efficiency of the Carnot's engine.

Ques. The Carnot heat engine operates on a

1. Reversible Carnot cycle
2. Non-reversible Carnot cycle

Ans. The correct answer a. Reversible Carnot cycle

Explanation: The Carnot heat engine is a theoretical engine that operates on a reversible Carnot cycle.

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Short Answers Questions [2 Marks Questions]

Ques. What is a Carnot engine?

Ans. A theoretical heat engine based on the Carnot cycle is known as a Carnot engine. Nicolas Léonard Sadi Carnot created the fundamental model for this engine in 1824. The Carnot engine could be the most efficient heat engine possible.

Ques. What is the Carnot theorem?

Ans. According to Carnot's theorem

• No heat engine has efficiency more than that of the Carnot engine.
• The efficiency of the Carnot engine is independent of the nature of the working substance.

Ques. What is the formula for the efficiency of a Carnot engine?

Ans. The formula for the efficiency of a Carnot engine is given by

µ = 1 - (T₂/T₁)

Where 

• T₁ is the source temperature
• T₂ is the sink temperature.

Ques. What is a Carnot cycle?

Ans. A Carnot cycle is an ideal thermodynamic cycle proposed by Sadi Carnot. It is a theoretical cycle that defines the maximum efficiency of a heat engine working between two different temperature heat reservoirs.

Ques. What is a heat engine?

Ans. Any device that transforms heat, partly into work or mechanical energy is called a heat engine. A quantity of matter inside the heat engine undergoes expansion or compression by absorbing or releasing heat is called the working substance of the heat engine.

Ques. What are the four processes involved in the reverse Carnot cycle?

Ans. The following are the four processes involved in the reverse Carnot cycle

• Isothermal expansion
• Adiabatic expansion
• Isothermal compression
• Adiabatic compression

Ques. On what factors does the efficiency of a Carnot engine depend?

Ans. The following are the factors on which the efficiency of a Carnot engine depends

• Temperature of the source
• Temperature of the sink

Ques. What is a reversible process?

Ans. The reversible process is a process in which at any stage of the process it can be traversed back in the opposite direction in such a way that the system passes through exactly the same conditions at every step in the reverse process as in the direct process.

Ques. What is an irreversible process?

Ans. An irreversible process may be defined as the process in which the system cannot be retraced to its initial state. Thermodynamic processes that occur in nature are all irreversible processes.

Ques. What is a thermodynamic system?

Ans. A collection of an extremely large number of atoms or molecules confined within certain boundaries such that it has a certain values of pressure (P), volume (V), and temperature (T) is called a thermodynamic system.

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Long Answers Questions [3 Marks Questions]

Ques. In a Carnot engine, when T₂ = 0 ℃ and T₁ = 200 ℃, its efficiency is µ₁ and when T₂ = 0 ℃ and T₁ = – 200 ℃, its efficiency is µ₂. What is the ratio of µ₁/µ₂?

Ans. The efficiency of the Carnot engine is given by

µ = 1 - (T₂/T₁)

Case 1: 

When T₂ = 0 ℃ = 273 K and T₁ = 200 ℃ = 473 K, then

µ₁ = 1 - (273/473) = (473 - 273)/473 = 200/473

Case 2: 

When T₂ = 0 ℃ = 273 K and T₁ = -200 ℃ = 73 K, then

µ₂ = 1 - (73/273) = (273 - 73)/273 = 200/273

Therefore

µ₁/µ₂ = (200/473) x (273/200) = 0.577

Ques. What are the applications of the Carnot engine?

Ans. The Carnot engine has practical uses in various thermal devices and machines, mainly because mechanical work requires a temperature gradient. Here are a few examples of uses, including real-world modeling.

• To evaluate practical engines in terms of their theoretical maximum efficiency, particularly during the design stage.
• To compare the efficiency of the actual Otto (petrol) and Diesel cycles as a theoretical reference cycle.
• To transform heat energy from the ocean into work.

The Carnot engine, on the other hand, is impossible because the heat transfer into the engine during the isothermal phase is too slow to be useful.

Ques. What are the properties of a Carnot engine? Why Carnot cycle is not practically possible

Ans. The following are the properties of a Carnot engine

• It has 100% efficiency.
• The pressure and volume of the working substance change slowly.
• The working substance is constantly in thermal equilibrium with the source and sink.
• As a working substance, a fully insulating frictionless piston and a cylinder with perfectly insulating walls are used; they are thought to have no dissipative effects like friction, heat conduction, and so on.

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Very Long Answers Questions [5 Marks Questions]

Ques. A Carnot engine has an efficiency of 1/5. The efficiency becomes 1/3 when the temperature of the sink is decreased by 50 K. What is the temperature of the sink?

Ans. The efficiency of the Carnot engine is given by

µ = 1 - (T₂/T₁)

Where 

• T₂ is the temperature of the sink
• T₁ is the temperature of the source

Let x be the temperature of the sink.

Case 1: When the temperature of the sink is T₂ = x, then the efficiency of the Carnot engine is 1/5.

⇒ 1/5 = 1 - (x/T₁)

⇒ T₁ = 5x/4

Case 2: When the temperature of the sink is T₂ = x - 50, then the efficiency of the Carnot engine is 1/3.

⇒ 1/3 = 1 - [(x - 50)/T₁]

⇒ x = (2T₁/3) + 50

On substituting the value of T₁ from case 1, we get

x = [2(5x/4)/3] + 50

⇒ 2x = 600

⇒ x = 300 K

Hence the temperature of the sink will be 300 K.

Ques. A Carnot engine has an efficiency of 60%. If the source is at 527 ℃, then find the temperature of the sink.

Ans. Given

• Temperature of source, T₁ = 527 ℃
• The efficiency of the Carnot engine, µ = 60% = 60/100 = 0.6

The temperature of the source in the Kelvin scale is given by

T₁ = 527 + 273 = 800 K

The efficiency of the Carnot engine is given by

µ = 1 - (T₂/T₁)

Where T₂ is the temperature of the sink

⇒ T₂ = (1 - µ)T₁

On substituting the values, we get

T₂ = (1 - 0.6)800 = 320 K

In Celcius scale, T₂ = 320 - 273 = 47 ℃

Ques. Find the efficiency of a Carnot engine whose source and sink are at 327 ℃ and 27 ℃.

Ans. Given

• The temperature of the source, T₁ = 327 ℃
• The temperature of the sink, T₂ = 27 ℃

In Kelvin scale

• The temperature of the source, T₁ = 327 ℃ + 273 = 600 K
• The temperature of the sink, T₂ = 27 ℃ + 273 = 300 K

The efficiency of the Carnot engine is given by

µ = 1 - (T₂/T₁)

On substituting the values, we get

µ = 1 - (300/600) = 0.5 or 50%

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