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Mechanical Engineering Exam  >  Mechanical Engineering Tests  >  Past Year Questions: Second Law Of Thermodynamics, Carnot Cycle And Entropy - Mechanical Engineering MCQ

Past Year Questions: Second Law Of Thermodynamics, Carnot Cycle And Entropy - Mechanical Engineering MCQ


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24 Questions MCQ Test - Past Year Questions: Second Law Of Thermodynamics, Carnot Cycle And Entropy

Past Year Questions: Second Law Of Thermodynamics, Carnot Cycle And Entropy for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Past Year Questions: Second Law Of Thermodynamics, Carnot Cycle And Entropy questions and answers have been prepared according to the Mechanical Engineering exam syllabus.The Past Year Questions: Second Law Of Thermodynamics, Carnot Cycle And Entropy MCQs are made for Mechanical Engineering 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Past Year Questions: Second Law Of Thermodynamics, Carnot Cycle And Entropy below.
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Past Year Questions: Second Law Of Thermodynamics, Carnot Cycle And Entropy - Question 1
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A condenser of a refrigeration system rejects heat at a rate of 120 kW, while is compressor consumes a power of 30 kW. The coefficient of performance of the system would be

[1992]

Detailed Solution for Past Year Questions: Second Law Of Thermodynamics, Carnot Cycle And Entropy - Question 1


Q1 = 120 kW
W = 30 × kW
Q2 = 120 – 30 = 90kW

Past Year Questions: Second Law Of Thermodynamics, Carnot Cycle And Entropy - Question 2
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A reversible heat transfer demands:

[1993]

Detailed Solution for Past Year Questions: Second Law Of Thermodynamics, Carnot Cycle And Entropy - Question 2

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Past Year Questions: Second Law Of Thermodynamics, Carnot Cycle And Entropy - Question 3
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When a system executes an irreversible cycle

[1994]

Detailed Solution for Past Year Questions: Second Law Of Thermodynamics, Carnot Cycle And Entropy - Question 3

Past Year Questions: Second Law Of Thermodynamics, Carnot Cycle And Entropy - Question 4
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A solar energy based heat engine which receives 80 kJ of heat at 100°C and rejects 70 kJ of heat to the ambient at 30°C is to be designed. The thermal efficiency of the heat engine is

[1996]
Detailed Solution for Past Year Questions: Second Law Of Thermodynamics, Carnot Cycle And Entropy - Question 4

Past Year Questions: Second Law Of Thermodynamics, Carnot Cycle And Entropy - Question 5
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For two cycles coupled in series, the topping cycle has an efficiency of 30% and the bottoming cycle has an efficiency of 20%. The overall combined cycle efficiency is

[1996]
Detailed Solution for Past Year Questions: Second Law Of Thermodynamics, Carnot Cycle And Entropy - Question 5

η0 = η1 + η2 - η1 η2
= 0.3 + 0.2 – (0.3) (0.2) = 0.44 = 44%

Past Year Questions: Second Law Of Thermodynamics, Carnot Cycle And Entropy - Question 6
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A cycle heat engine does 50 kJ of work per cycle. If the efficiency of the heat engine is 75%. The heat rejected per cycle is

[2001]
Detailed Solution for Past Year Questions: Second Law Of Thermodynamics, Carnot Cycle And Entropy - Question 6

h = 0.75

or, Net heat input,

∴ Net heat required, Q2 = 50/3 = 16.66kJ

Past Year Questions: Second Law Of Thermodynamics, Carnot Cycle And Entropy - Question 7
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A solar collector receiving solar radiation at the rate of 0.6 kW/m2 transforms it to the internal energy of a fluid at an overall efficiency of 50%. The fluid heated to 350 K is used to run a heat engine which rejects heat at 313 K.If the heat engine is to deliver 2.5 kW power, the minimum area of the solar collector required would be

[2004]
Detailed Solution for Past Year Questions: Second Law Of Thermodynamics, Carnot Cycle And Entropy - Question 7

Given; Receiving solar radiation at the rate of 0.6 kW/m2
Internal energy of fluid after absorbing solar radiation


Let A be minimum area of collector

or 25 kW = 0.3 kW/m2
or A = 25/0.3 = 83.33 m2

Past Year Questions: Second Law Of Thermodynamics, Carnot Cycle And Entropy - Question 8
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A heat transformer is device that transfers a part of the heat, supplied to it at an intermediate temperature, to a high temperature reservoir while rejecting the remaining part to a low temperature heat sink.In such a heat transformer, 100 kJ of heat is supplied at 350 K. The maximum amount of heat in kJ that can be transferred to 400 K, when the rest is rejected to a heat sink at 300 K is

[2007]
Detailed Solution for Past Year Questions: Second Law Of Thermodynamics, Carnot Cycle And Entropy - Question 8


Past Year Questions: Second Law Of Thermodynamics, Carnot Cycle And Entropy - Question 9
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A cyclic device operates between three thermal reservoirs, as shown in the figure. Heat is transferred to/from the cycle device. It is assumed that heat transfer between each thermal reservoir and the cyclic device takes place across negligible temperature difference.
Interactions between the cyclic device and the respective thermal reservoirs that are shown in the figure are all in the form of heat transfer.

The cyclic device can be

[2008]
Detailed Solution for Past Year Questions: Second Law Of Thermodynamics, Carnot Cycle And Entropy - Question 9

Since heat is taken from the high temperature sources and rejected to low temperature sink, hence the device is a heat engine not heat pump.
Since, the temperature differences ar e negligible, the engine is reversible.

Past Year Questions: Second Law Of Thermodynamics, Carnot Cycle And Entropy - Question 10
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An irreversible heat engine extracts heat from a high temperature source at a rate of 100 kW and rejects heat to a sink at a rate of 50 kW.The entire work output of the heat engine is used to drive a reversible heat pump operating between a set of independent isothermal heat reservoirs at 17°C and 75°C. The rate (in kW) at which the heat pump delivers heat to its high temperature sink is

[2009]
Detailed Solution for Past Year Questions: Second Law Of Thermodynamics, Carnot Cycle And Entropy - Question 10


W = Q1 – Q2
= 100 – 50
= 50 kW

Q1 = 300 kW
Alternately
Work output from irreversible heat engine,
W = Q1 – Q2
= 100 –50 = 50 kW

Also for irreversible heat engine
Q1 + W = Q2
∴ T1 ΔS + 50 = T2ΔS
or (T2 – T1) ΔS = 50
or (348 - 290) Δ S = 50
or ΔS = 50/58
∴ Rate of heat rejection,
Q2 = T2 ΔS

Past Year Questions: Second Law Of Thermodynamics, Carnot Cycle And Entropy - Question 11
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Consider the following two processes;
I. A heat source at 1200 K loses 2500 kJ of heat to a sink at 800 K
II. A heat source at 800 K loses 2000 kJ of heat to a sink at 500 K
Which of the following statements is true?

[2010]
Detailed Solution for Past Year Questions: Second Law Of Thermodynamics, Carnot Cycle And Entropy - Question 11

From Clausius inequality,

For process I


For process II


Hence process II is more irreversible than process I.

Past Year Questions: Second Law Of Thermodynamics, Carnot Cycle And Entropy - Question 12
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A reversible heat engine receives 2 kJ of heat from a reservoir at 1000 K and a certain amount of heat from a reservoir at 800 K.It rejects 1 kJ of heat to a reservoir at 400 K.
The net work output (in kJ) of the cycle is

[2014, Set-1]
Detailed Solution for Past Year Questions: Second Law Of Thermodynamics, Carnot Cycle And Entropy - Question 12


We know that for reversible heat engine, change in entropy is always zero
That is ΔS = 0

Q2 = 0.4 kJ
WNet = (Q1 + Q2) – Q3 = (2 + 0.4) – 1 = 1.4 kJ.

Past Year Questions: Second Law Of Thermodynamics, Carnot Cycle And Entropy - Question 13
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The figure shows a heat engine (HE) working between two reservoirs. The amount of heat (Q2) rejected by the heat engine is drawn by a heat pump (HP). The heat pump receives the entire work out put (W) of the heat engine. If temperatures, T1 > T3 > T2, then the relation between the efficiency (η) of the heat engine and the coefficient and the coefficient of performance (COP) of the heat pump is

[2019, Set -2]
Detailed Solution for Past Year Questions: Second Law Of Thermodynamics, Carnot Cycle And Entropy - Question 13

Efficiency of heat engine,

∵ Q3 = Q2 + W = Q2 + Q1 – Q2
Q3 = Q1

Past Year Questions: Second Law Of Thermodynamics, Carnot Cycle And Entropy - Question 14
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Consider a refrigerator and a heat pump working on the reversed Carnot cycle between the same temperature limits. Which of the following is correct?

[1995]
Past Year Questions: Second Law Of Thermodynamics, Carnot Cycle And Entropy - Question 15
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A Carnot cycle is having an efficiency of 0.75. If the temperature of the high temperature reservoir is 727°C. What is the temperature of low temperature reservoir?

[2001]
Detailed Solution for Past Year Questions: Second Law Of Thermodynamics, Carnot Cycle And Entropy - Question 15


T2 = 250K
T2 = 273 – 250
[T2 – 23°C]

Past Year Questions: Second Law Of Thermodynamics, Carnot Cycle And Entropy - Question 16
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Which among the following relations is/are valid only for reversible process undergone by a pure substance?

[1993]
Past Year Questions: Second Law Of Thermodynamics, Carnot Cycle And Entropy - Question 17
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One kilomole of an ideal gas is throttled from an initial pressure of 0.5 MPa to 0.1 MPa. The initial temperature is 300 K. The entropy change of the universe is

[1995]
Detailed Solution for Past Year Questions: Second Law Of Thermodynamics, Carnot Cycle And Entropy - Question 17

Throttling, Ti = Tf
Hence

8.314 ln 5
(ds) = 13.388 kJ/K

Past Year Questions: Second Law Of Thermodynamics, Carnot Cycle And Entropy - Question 18
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For an ideal gas the expression


is always equal to

[1997]
Past Year Questions: Second Law Of Thermodynamics, Carnot Cycle And Entropy - Question 19
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A system undergoes a State change from 1 to 2. According to the second law of thermodynamics, for the process to be feasible, the entropy change, S2 – S1 of the system

[1997]
Detailed Solution for Past Year Questions: Second Law Of Thermodynamics, Carnot Cycle And Entropy - Question 19

Entropy of irreversib le process alw ays increases.

Past Year Questions: Second Law Of Thermodynamics, Carnot Cycle And Entropy - Question 20
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If a closed system is undergoing an irreversible process, the entropy of the system

[2009]
Detailed Solution for Past Year Questions: Second Law Of Thermodynamics, Carnot Cycle And Entropy - Question 20

(ds)uni ≥ 0

Past Year Questions: Second Law Of Thermodynamics, Carnot Cycle And Entropy - Question 21
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One kilogram of water at room temperature is brought into contact with a high temperature thermal reservoir. The entropy change of the universe is

[2010]
Detailed Solution for Past Year Questions: Second Law Of Thermodynamics, Carnot Cycle And Entropy - Question 21

Explanation:

Reversible process: A thermodynamic process driving from initial state to final state is said to be reversible, if the system as well as its surrounding returns back to its initial state, without any change in the universe.

Irreversible process: It is a process that does not return back to its initial state. 

Change in Entropy of the system can be expressed as shown below 

From the above explanation, we can see that the equality sign holds good when the process undergone by the system is reversible, the inequality sign holds good if there is any irreversibility present in the process. This statement is usually called the principle of entropy increase.

For an isolated system,

  • ΔS > 0, for irreversible processes.
  • ΔS = 0, for reversible processes
  • ΔS < 0, the process is impossible.

However, for the reverse process, ΔS = 0 (i.e. Entropy remains constant.) ​.

In the question, it is not mentioned whether the process is reversible or irreversible. Therefore here we will take a general process i.e irreversible. For an irreversible process, the change in the entropy of the universe will be always positive. 

Past Year Questions: Second Law Of Thermodynamics, Carnot Cycle And Entropy - Question 22
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An ideal gas of mass m and temperature T1 undergoes a reversible isothermal process from an initial pressure P1 to final pressure P2. The heat loss during the process is Q. The entropy change ΔS of the gas is

[2012]
Past Year Questions: Second Law Of Thermodynamics, Carnot Cycle And Entropy - Question 23
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Which one of the following pairs of equations describes an irreversible heat engine?

[2014, Set-3]
Past Year Questions: Second Law Of Thermodynamics, Carnot Cycle And Entropy - Question 24
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For an ideal gas with constant properties undergoing a quasi-static process, which one of the following represents the change of entropy (Δs) from state 1 to 2?

[2018, Set-2]
Detailed Solution for Past Year Questions: Second Law Of Thermodynamics, Carnot Cycle And Entropy - Question 24


Tds = dh – vdp
= Cpdt – vdp

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