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Test: Second Law of Thermodynamics - 1 - Mechanical Engineering MCQ


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15 Questions MCQ Test Thermodynamics - Test: Second Law of Thermodynamics - 1

Test: Second Law of Thermodynamics - 1 for Mechanical Engineering 2024 is part of Thermodynamics preparation. The Test: Second Law of Thermodynamics - 1 questions and answers have been prepared according to the Mechanical Engineering exam syllabus.The Test: Second Law of Thermodynamics - 1 MCQs are made for Mechanical Engineering 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Second Law of Thermodynamics - 1 below.
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Test: Second Law of Thermodynamics - 1 - Question 1

Select the Kelvin-Plank statement of the second law:

Test: Second Law of Thermodynamics - 1 - Question 2

According to the Clausius statement of the second law:
1. heat flows from cold surface to hot surface, unaided.
2. heat flows from hot surface to cold surface, unaided.
3. heat can flow from cold surface to hot surface with the aid of external work

Which of the above statements is/are correct?

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Test: Second Law of Thermodynamics - 1 - Question 3

If a heat engine attains 100% thermal efficiency, it violates

Detailed Solution for Test: Second Law of Thermodynamics - 1 - Question 3

Such a heat engine is PMM2 which is impossible. It violates kelvin-Plank statement

Test: Second Law of Thermodynamics - 1 - Question 4

A refrigerator and a heat pump operate between the same temperature limits. If the COP of the refrigerator is 4, the COP of the heat pump would be

Detailed Solution for Test: Second Law of Thermodynamics - 1 - Question 4

Test: Second Law of Thermodynamics - 1 - Question 5

A condenser of a refrigeration system rejects heat at a rate of 120 kW, while its compressor consumes a power of 30 kW. The coefficient of performance of the system would be

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Test: Second Law of Thermodynamics - 1 - Question 6

If the temperature of the source is increased keeping sink temperature fixed, the efficiency of the Carnot engine

Detailed Solution for Test: Second Law of Thermodynamics - 1 - Question 6

Test: Second Law of Thermodynamics - 1 - Question 7

An industrial heat pump operates between the temperatures of 27°C and –13°C. The rates of heat addition and heat rejection are 750 W and 1000 W, respectively. The COP for the heat pump is

Detailed Solution for Test: Second Law of Thermodynamics - 1 - Question 7

It may not be a reversible heat pump. So we use only heat data to get COP

Hence, the correct option is (c).

Test: Second Law of Thermodynamics - 1 - Question 8

The more effective way of increasing the efficiency of a Carnot engine is to

Detailed Solution for Test: Second Law of Thermodynamics - 1 - Question 8

(decreasing lower temperature by ΔT)

(increasing higher temperature by ΔT)

So, the more effective way to increase the cycle efficiency is to decrease lower temperature.

Test: Second Law of Thermodynamics - 1 - Question 9

A Carnot engine operates between reservoirs at 20°C and 200°C. If 10 kW of power is produced, the rejected heat rate is nearest

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Test: Second Law of Thermodynamics - 1 - Question 10

An inventor invents a thermal engine that operates between ocean layers at 27°C and 10°C. It produces 10 kW and discharges 9900 kJ/min of heat. Such an enigne is

Detailed Solution for Test: Second Law of Thermodynamics - 1 - Question 10

Actual efficiency is more than the 2nd law efficiency which is impossible.

Test: Second Law of Thermodynamics - 1 - Question 11

If the time taken by a system to execute a process through a finite gradient is infinitely large, the process

Test: Second Law of Thermodynamics - 1 - Question 12

“Heat cannot be transported from a system at low temperature to another system at high temperature without the aid of external agency”. This statement of second law is attributed to

*Answer can only contain numeric values
Test: Second Law of Thermodynamics - 1 - Question 13

The heat removal rate from a refrigerated space and the power input to the compressor are 7.2 kW and 1.8 kW, respectively. The coefficient of performance (COP) of the refrigerator is .


Detailed Solution for Test: Second Law of Thermodynamics - 1 - Question 13

Heat removal rate, Q2 = 7.2 kW

Power input to the compressor W = 1.8 kW

The coefficient of performance (COP) of the refrigerator can be calculated using the relation 

Hence the correct answer is 4.0

Test: Second Law of Thermodynamics - 1 - Question 14

Any thermodynamic cycle operating between two temperature limits is reversible if the product of the efficiency when operating as a heat engine and the COP when operating as a refrigerator is equal to 1.

Detailed Solution for Test: Second Law of Thermodynamics - 1 - Question 14

For reversible engine,

 

For reversible refrigerator

.

Hence, product for reversible thermodynamic cycle is not equal to 1.

*Answer can only contain numeric values
Test: Second Law of Thermodynamics - 1 - Question 15

A Carnot engine (CE-1) works between two temperature reservoirs A and B, where TA = 900 K and TB = 500 K. A second Carnot engine (CE-2) works between temperature reservoirs B and C, where TC = 300 K. In each cycle CE-1 and CE-2, all the heat rejected by CE-1 to reservoir B is used by CE-2. For one cycle of operation, if the net Q absorbed by CE-1 from reservoir A is 150 MJ, the net heat rejected to reservoir C by CE-2 (in MJ) is .


Detailed Solution for Test: Second Law of Thermodynamics - 1 - Question 15

For CE1:

For CE2:

Hence, the correct answer is 50.

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