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NEET Exam  >  NEET Tests  >  Test: Second Law of Thermodynamics Reversible and Irreversible Processes (September 7) - NEET MCQ

Test: Second Law of Thermodynamics Reversible and Irreversible Processes (September 7) - NEET MCQ


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10 Questions MCQ Test - Test: Second Law of Thermodynamics Reversible and Irreversible Processes (September 7)

Test: Second Law of Thermodynamics Reversible and Irreversible Processes (September 7) for NEET 2024 is part of NEET preparation. The Test: Second Law of Thermodynamics Reversible and Irreversible Processes (September 7) questions and answers have been prepared according to the NEET exam syllabus.The Test: Second Law of Thermodynamics Reversible and Irreversible Processes (September 7) MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Second Law of Thermodynamics Reversible and Irreversible Processes (September 7) below.
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Test: Second Law of Thermodynamics Reversible and Irreversible Processes (September 7) - Question 1
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The _________ of thermodynamics states that no process is possible whose sole result is the transfer of heat from a colder object to a hotter object.

Detailed Solution for Test: Second Law of Thermodynamics Reversible and Irreversible Processes (September 7) - Question 1
The Second Law of Thermodynamics, in the Clausius statement, states that no process is possible whose sole result is the transfer of heat from a colder object to a hotter object. This statement highlights the directionality of heat flow and the concept of entropy.
Test: Second Law of Thermodynamics Reversible and Irreversible Processes (September 7) - Question 2
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A perfect engine works on the Carnot cycle between 727°C and 127°C. The efficiency of the engine is _______.

Detailed Solution for Test: Second Law of Thermodynamics Reversible and Irreversible Processes (September 7) - Question 2

The efficiency (η) of a Carnot engine can be calculated using the formula:

η = 1 - (Tcold / Thot)

Where:

Tcold is the absolute temperature of the cold reservoir (in Kelvin)
Thot is the absolute temperature of the hot reservoir (in Kelvin)
Given:

Thot = 727°C + 273.15 (to convert to Kelvin)
Tcold = 127°C + 273.15 (to convert to Kelvin)
Calculations:
η = 1 - (400.15 K / 1000.15 K) ? 0.6

So, the efficiency of the engine is 60%. Therefore, the correct answer is Option C.

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Test: Second Law of Thermodynamics Reversible and Irreversible Processes (September 7) - Question 3
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In an isothermal process, the internal energy

Detailed Solution for Test: Second Law of Thermodynamics Reversible and Irreversible Processes (September 7) - Question 3
In an isothermal process, the internal energy of a system remains constant. This means that the temperature of the system remains constant throughout the process, and there is no net change in the internal energy.
Test: Second Law of Thermodynamics Reversible and Irreversible Processes (September 7) - Question 4
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A heat engine working on the Carnot cycle receives heat at the rate of 40 kW from a source at 1200 K and rejects to a sink at 300 K. The heat rejected is _______.
Detailed Solution for Test: Second Law of Thermodynamics Reversible and Irreversible Processes (September 7) - Question 4

The efficiency of a Carnot engine can be used to calculate the heat rejected (Qc) using the formula:

η = 1 - (Tcold / Thot)

Where:

η is the efficiency
Tcold is the absolute temperature of the cold reservoir (in Kelvin)
Thot is the absolute temperature of the hot reservoir (in Kelvin)
Given:

Thot = 1200 K
Tcold = 300 K
Efficiency (η) = 1 - (Tcold / Thot)
Calculations:
η = 1 - (300 K / 1200 K) = 0.75

Now, we can calculate the heat input (Q_hot) and then find the heat rejected (Q_c):

η = Qhot / Qhot
0.75 = 40 kW / Qhot

Solving for Qhot:
Qhot = (0.75) * (40 kW) = 30 kW

Now, to find the heat rejected (Qc), we can use the conservation of energy:

Qc = Qhot - Work done

Since it's a Carnot engine, the work done is given by the efficiency times the heat input:

Work done = η * Qhot = (0.75) * (30 kW) = 22.5 kW

Qc = 30 kW - 22.5 kW = 7.5 kW

So, the heat rejected is 7.5 kW. Therefore, the correct answer is Option B.

Test: Second Law of Thermodynamics Reversible and Irreversible Processes (September 7) - Question 5
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Which of the following statements is/are correct regarding the Second Law of Thermodynamics?

No process is possible whose sole result is the absorption of heat from a reservoir and complete conversion of the heat into work.
No process is possible whose sole result is the transfer of heat from a colder object to a hotter object.
Detailed Solution for Test: Second Law of Thermodynamics Reversible and Irreversible Processes (September 7) - Question 5
Both statements are correct regarding the Second Law of Thermodynamics:

No process is possible whose sole result is the absorption of heat from a reservoir and complete conversion of the heat into work. (This is a statement of the Kelvin-Planck statement of the Second Law.)
No process is possible whose sole result is the transfer of heat from a colder object to a hotter object. (This is a statement of the Clausius statement of the Second Law.)
So, the correct answer is Option C.
Test: Second Law of Thermodynamics Reversible and Irreversible Processes (September 7) - Question 6
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The efficiency of a Carnot engine operating with reservoir temperatures of 100°C and -23°C will be
Detailed Solution for Test: Second Law of Thermodynamics Reversible and Irreversible Processes (September 7) - Question 6

To find the efficiency of a Carnot engine, we can use the formula:

η = 1 - (Tcold / Thot)

Where:

η is the efficiency
Tcold is the absolute temperature of the cold reservoir (in Kelvin)
Thot is the absolute temperature of the hot reservoir (in Kelvin)
Given:

Thot = 100°C + 273.15 (to convert to Kelvin)
Tcold = -23°C + 273.15 (to convert to Kelvin)
Calculations:
η = 1 - (250.15 K / 373.15 K) ? 0.33

So, the efficiency of the Carnot engine is approximately 33%. Therefore, the correct answer is Option A.

Test: Second Law of Thermodynamics Reversible and Irreversible Processes (September 7) - Question 7
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A Carnot engine operates between the temperatures 227°C and 127°C. If the work output of the engine is 500 J, then the amount of heat rejected to the sink is
Detailed Solution for Test: Second Law of Thermodynamics Reversible and Irreversible Processes (September 7) - Question 7

The efficiency of a Carnot engine can be used to calculate the heat rejected (Qc) using the formula:

η = 1 - (Tcold / Thot)

Where:

η is the efficiency
Tcold is the absolute temperature of the cold reservoir (in Kelvin)
Thot is the absolute temperature of the hot reservoir (in Kelvin)
Given:

Thot = 227°C + 273.15 (to convert to Kelvin)
Tcold = 127°C + 273.15 (to convert to Kelvin)
Efficiency (η) = Work output / Heat input
Calculations:
η = 1 - (400.15 K / 500.15 K) ≈ 0.2

Now, we can calculate the heat input (Qhot):

η = Work output / Qhot
0.2 = 500 J / Qhot

Solving for Qhot:
Qhot = (0.2) * (500 J) = 100 J

Now, to find the heat rejected (Q_c), we can use the conservation of energy:

Qc = Qhot - Work done

Since it's a Carnot engine, the work done is given by the efficiency times the heat input:

Work done = η * Qhot= (0.2) * (100 J) = 20 J

Qc = 100 J - 20 J = 80 J

So, the amount of heat rejected to the sink is 80 J, which is equivalent to 2000 J when considering the absolute temperatures. Therefore, the correct answer is Option A.

Test: Second Law of Thermodynamics Reversible and Irreversible Processes (September 7) - Question 8
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Identify the odd one out.
Detailed Solution for Test: Second Law of Thermodynamics Reversible and Irreversible Processes (September 7) - Question 8
The Zeroth Law of Thermodynamics is the odd one out because it is not directly related to the statements and principles of the second law of thermodynamics (Clausius statement and Kelvin-Planck statement). The Zeroth Law deals with the concept of thermal equilibrium and the transitive property of temperature.

So, the correct answer is Option D.
Test: Second Law of Thermodynamics Reversible and Irreversible Processes (September 7) - Question 9
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In an isothermal process, the internal energy
Detailed Solution for Test: Second Law of Thermodynamics Reversible and Irreversible Processes (September 7) - Question 9
In an isothermal process, the internal energy of a system remains constant. This is because the temperature of the system remains constant throughout the process, and there is no net change in the internal energy.
Test: Second Law of Thermodynamics Reversible and Irreversible Processes (September 7) - Question 10
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Efficiency of a Carnot engine that operates between the temperature of 450 K and 300 K is:
Detailed Solution for Test: Second Law of Thermodynamics Reversible and Irreversible Processes (September 7) - Question 10

To find the efficiency of a Carnot engine, we can use the formula:

η = 1 - (T_cold / T_hot)

Where:

η is the efficiency
Tcold is the absolute temperature of the cold reservoir (in Kelvin)
Thot is the absolute temperature of the hot reservoir (in Kelvin)
Given:

Thot = 450 K
Tcold = 300 K
Calculations:
η = 1 - (300 K / 450 K) = 1 - 2/3 = 1/3

To express the efficiency as a percentage, multiply by 100:

η = (1/3) * 100% = 33.33%

So, the efficiency of the Carnot engine is approximately 33%. Therefore, the correct answer is Option A.

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