100 High-Scoring MCQs from Heat and Thermodynamics | Physics Class 11 | Unit 10

100 Important MCQs on Heat and Thermodynamics, Chapter 10, Class 11 Physics (Unit-Wise Practice):

This post features 100 carefully selected MCQs from Heat and Thermodynamics (Unit 10) for Class 11 Physics. Each question is designed according to the latest syllabus and emphasizes:

• Concept clarity – strengthen your understanding of heat, work, and thermodynamic principles
• Numerical practice – solve important numerical problems with ease
• Exam preparation – perfect for FBISE, board, and competitive exam

Boost your knowledge, practice effectively, and enhance your exam readiness with this comprehensive MCQ collection!

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MCQs No.1

The first law of thermodynamics is based on the law of conservation of:
a. Mass                b. Momentum
c. Energy              d. Entropy

The Correct Answer is option c. Energy

Explanation:
The first law states that energy can neither be created nor destroyed, only transformed.

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MCQs No.2

Which of the following is a state function?
a. Heat                    b. Work
c. Temperature        d. Path length

The Correct Answer is option c. Temperature

Explanation:
Temperature depends only on the state of the system, not the path.

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MCQs No.3

The unit of entropy is:
a. Joule                    b. Joule per kelvin
c. Kelvin                    d. Joule per second

The Correct Answer is option b. Joule per kelvin

Explanation:

Entropy is defined as heat divided by temperature.

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MCQs No.4

An isothermal process occurs at constant:
a. Pressure                  b. Volume
c. Temperature            d. Entropy

The Correct Answer is option c. Temperature

Explanation:

In an isothermal process, temperature remains constant.

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MCQs No.5

In an adiabatic process, which quantity remains constant?
a. Heat                        b. Temperature
c. Entropy                    d. Pressure

The Correct Answer is option a. Heat

Explanation:

In an adiabatic process, no heat is exchanged with surroundings.

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MCQs No.6 (Numerical)

If 500 J of heat is supplied to a gas and it does 200 J of work, the change in internal energy is:
a. 300 J                b. 700 J
c. −300 J              d. −700 J

The Correct Answer is option a. 300 J

Explanation:

$$\Delta U=Q-W=500-200=300\text{J}$$

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MCQs No.7

Which law introduces the concept of entropy?
a. Zeroth law                b. First law
c. Second law                d. Third law

The Correct Answer is option c. Second law

Explanation:
The second law defines entropy and its natural tendency to increase.

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MCQs No.8

Entropy of the universe in a reversible process is:
a. Increased                b. Decreased
c. Infinite                    d. Zero

The Correct Answer is option d. Zero

Explanation:
In a reversible process, total entropy change is zero.

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MCQs No.9

A perpetual motion machine of first kind violates:
a. Zeroth law             b. First law
c. Second law            d. Third law

The Correct Answer is option b. First law

Explanation:
Such a machine creates energy without input, violating energy conservation.

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MCQs No.10

A perpetual motion machine of second kind violates:
a. Zeroth law                b. First law
c. Second law                d. Third law

The Correct Answer is option c. Second law

Explanation:
It converts heat completely into work, violating entropy law.

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MCQs No.11

Internal energy of an ideal gas depends on:
a. Pressure only                    b. Volume only
c. Temperature only              d. Pressure and volume

The Correct Answer is option c. Temperature only

Explanation:
For an ideal gas, internal energy is a function of temperature alone.

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MCQs No.12 (Numerical)

If 1000 J of heat is absorbed by a system and 400 J of work is done by it, the change in internal energy is:
a. 600 J                    b. 1400 J
c. −600 J                  d. −1400 J

The Correct Answer is option a. 600 J

Explanation:

$$\Delta U=Q-W=1000-400=600\text{J}$$

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MCQs No.13

In which process is work done maximum?
a. Isothermal                    b. Isochoric
c. Isobaric                        d. Adiabatic

The Correct Answer is option a. Isothermal

Explanation:
For given conditions, isothermal expansion does maximum work.

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MCQs No.14

In an isochoric process, work done is:
a. Maximum                    b. Minimum
c. Zero                            d. Infinite

The Correct Answer is option c. Zero

Explanation:
Volume remains constant, so no work is done.

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MCQs No.15

Which process is represented by a vertical line on a P–V diagram?
a. Isothermal                    b. Isobaric
c. Isochoric                      d. Adiabatic

The Correct Answer is option c. Isochoric

Explanation:
Constant volume is shown by a vertical line.

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MCQs No.16 (Numerical)

If a gas absorbs 300 J of heat and does no work, the change in internal energy is:
a. 0 J                            b. 300 J
c. −300 J                       d. 600 J

The Correct Answer is option b. 300 J

Explanation:

ΔU=Q=300J

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MCQs No.17

Heat and work are:
a. State functions
b. Path functions
c. Constant quantities
d. Vector quantities

The Correct Answer is option b. Path functions

Explanation:
Heat and work depend on the path followed.

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MCQs No.18

Which of the following is not a thermodynamic process?
a. Isothermal                    b. Isobaric
c. Isochoric                      d. Uniform motion

The Correct Answer is option d. Uniform motion

Explanation:
Uniform motion is not a thermodynamic process.

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MCQs No.19

The efficiency of a heat engine can never be:
a. Less than 1                    b. Equal to 1
c. Greater than 1                d. Zero

The Correct Answer is option c. Greater than 1

Explanation:
Efficiency cannot exceed 100%.

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MCQs No.20 (Numerical)

If a heat engine absorbs 500 J of heat and rejects 300 J, the work done is:
a. 200 J                    b. 800 J
c. −200 J                  d. 0 J

The Correct Answer is option a. 200 J

Explanation:

W=Qh​−Qc​=500−300=200J

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MCQs No.21

Which law states that heat cannot flow from a colder body to a hotter body without external work?
a. Zeroth law                b. First law
c. Second law                d. Third law

The Correct Answer is option c. Second law

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MCQs No.22

Entropy of an isolated system always:
a. Decreases                    b. Remains constant
c. Increases                     d. Becomes zero

The Correct Answer is option c. Increases

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MCQs No.23 (Numerical)

If the efficiency of a heat engine is 40% and heat absorbed is 1000 J, the work done is:
a. 400 J
b. 600 J
c. 1000 J
d. 1400 J

The Correct Answer is option a. 400 J

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MCQs No.24

Which process has no heat exchange?
a. Isothermal
b. Isobaric
c. Isochoric
d. Adiabatic

The Correct Answer is option d. Adiabatic

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MCQs No.25

The zeroth law of thermodynamics deals with:
a. Heat flow
b. Energy conservation
c. Thermal equilibrium
d. Entropy

The Correct Answer is option c. Thermal equilibrium

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MCQs No.26 (Numerical)

A system absorbs 600 J of heat and does 250 J of work. The change in internal energy is:
a. 350 J
b. 850 J
c. 600 J
d. 250 J

The Correct Answer is option a. 350 J

Explanation:

ΔU=Q−W=600−250=350J

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MCQs No.27

Which process occurs at constant pressure?
a. Isothermal
b. Isobaric
c. Isochoric
d. Adiabatic

The Correct Answer is option b. Isobaric

Explanation:
In an isobaric process, the pressure remains constant while volume or temperature may change.

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MCQs No.28 (Numerical)

A gas expands isothermally at 300 K, absorbing 400 J of heat. The work done by the gas is:
a. 400 J
b. 300 J
c. 100 J
d. 0 J

The Correct Answer is option a. 400 J

Explanation:
For an isothermal process of an ideal gas, $\Delta U=0$, so $Q=W=400\text{J}$.

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MCQs No.29

The second law of thermodynamics states that:
a. Energy can be created
b. Total entropy of universe increases in irreversible processes
c. Work is always zero
d. Temperature remains constant

The Correct Answer is option b. Total entropy of universe increases in irreversible processes

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MCQs No.30

An adiabatic process is represented by which type of curve on a P–V diagram?
a. Straight horizontal line
b. Vertical line
c. Hyperbolic curve steeper than isothermal
d. Circle

The Correct Answer is option c. Hyperbolic curve steeper than isothermal

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MCQs No.31 (Numerical)

A gas has an internal energy increase of 500 J when 800 J of heat is supplied. The work done by the gas is:
a. 300 J
b. 500 J
c. 800 J
d. 1300 J

The Correct Answer is option a. 300 J

Explanation:

$$W=Q-\Delta U=800-500=300\text{J}$$

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MCQs No.32

Which of the following is NOT a state function?
a. Internal energy
b. Entropy
c. Heat
d. Temperature

The Correct Answer is option c. Heat

Explanation:
Heat depends on the path taken, so it is a path function, not a state function.

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MCQs No.33

The efficiency of a Carnot engine depends on:
a. Type of working substance
b. Temperatures of heat reservoirs
c. Volume of the gas
d. Pressure of the gas

The Correct Answer is option b. Temperatures of heat reservoirs

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MCQs No.34 (Numerical)

A Carnot engine absorbs 1000 J of heat from a reservoir at 500 K and rejects heat to a reservoir at 300 K. The maximum efficiency is:
a. 0.4
b. 0.3
c. 0.6
d. 0.5

The Correct Answer is option a. 0.4

Explanation:

$$\eta = 1 - \frac{T_c}{T_h} = 1 - \frac{300}{500} = 0.4$$

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MCQs No.35 (Numerical)

If 200 J of work is done by a system and its internal energy decreases by 50 J, the heat absorbed is:
a. 150 J
b. 250 J
c. 50 J
d. 200 J

The Correct Answer is option a. 150 J

Explanation:

$$Q=\Delta U+W=(-50)+200=150\text{J}$$

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MCQs No.36

Which of the following is true for an isolated system?
a. $\Delta U = 0$
b.  $Q=0$ and  $W = 0$
c.  $\Delta S \ge 0$
d. All of the above

The Correct Answer is option d. All of the above

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MCQs No.37

Entropy of a reversible adiabatic process is:
a. Increased
b. Decreased
c. Zero
d. Infinite

The Correct Answer is option c. Zero

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MCQs No.38 (Numerical)

A system receives 400 J of heat and does 150 J of work. What is the increase in internal energy?
a. 250 J
b. 400 J
c. 550 J
d. 150 J

The Correct Answer is option a. 250 J

Explanation:

$$\Delta U=Q-W=400-150=250\text{J}$$

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MCQs No.39

Which of the following processes is isochoric?
a. Constant pressure
b. Constant temperature
c. Constant volume
d. Constant entropy

The Correct Answer is option c. Constant volume

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MCQs No.40 (Numerical)

A heat engine absorbs 500 J of heat and rejects 300 J. What is the efficiency?
a. 0.2
b. 0.4
c. 0.5
d. 0.6

The Correct Answer is option b. 0.4

Explanation:

$$\eta = \frac{W}{Q_h} = \frac{Q_h - Q_c}{Q_h} = \frac{500-300}{500} = 0.4$$

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MCQs No.41

Work done in an isothermal expansion of ideal gas is:
a. Zero
b. Maximum
c. Minimum
d. Infinite

The Correct Answer is option b. Maximum

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MCQs No.42

For a reversible adiabatic process:
a. $\Delta U=0$
b. $Q=0$
c. $W = 0$
d. $\Delta S>0$

The Correct Answer is option b. Q = 0

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MCQs No.43 (Numerical)

A system does 250 J of work and loses 50 J of internal energy. How much heat is exchanged?
a. 300 J absorbed
b. 200 J released
c. 250 J absorbed
d. 300 J released

The Correct Answer is option b. 200 J absorbed

Explanation:

$$Q = \Delta U + W = -50 + 250 = 200 \, \text{J}$$

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MCQs No.44

Zeroth law of thermodynamics is used to define:
a. Heat
b. Work
c. Temperature
d. Entropy

The Correct Answer is option c. Temperature

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MCQs No.45 (Numerical)

If 600 J of heat is absorbed in an adiabatic process, the work done by the gas is:
a. 0 J
b. 300 J
c. 600 J
d. 100 J

The Correct Answer is option a. 0 J

Explanation:
In adiabatic process, $Q=0$. If heat is absorbed by system → process not adiabatic.

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MCQs No.46

Entropy increases in:
a. Reversible process
b. Irreversible process
c. Both
d. Neither

The Correct Answer is option b. Irreversible process

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MCQs No.47 (Numerical)

A gas expands from 2 L to 4 L at 300 K isothermally. If $nR = 8.314$, work done is:
a. 2.08 kJ
b. 1.73 kJ
c. 4.16 kJ
d. 0.5 kJ

The Correct Answer is option b. 1.73 kJ

Explanation:

$$W = nRT \ln \frac{V_f}{V_i} = 8.314 \times 300 \ln \frac{4}{2} \approx 1.73 \, \text{kJ}$$

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MCQs No.48

Which of the following is an extensive property?
a. Temperature
b. Pressure
c. Volume
d. Refractive index

The Correct Answer is option c. Volume

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MCQs No.49 (Numerical)

An engine absorbs 1200 J of heat and does 480 J of work. The efficiency is:
a. 0.2
b. 0.3
c. 0.4
d. 0.5

The Correct Answer is option c. 0.4

Explanation:

$$\eta = \frac{W}{Q_h} = \frac{480}{1200} = 0.4$$

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MCQs No.50

The first law of thermodynamics is a statement of:
a. Conservation of momentum
b. Conservation of energy
c. Conservation of mass
d. Conservation of entropy

The Correct Answer is option b. Conservation of energy

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MCQs No.51

The SI unit of heat is:
a. Calorie        b. Joule        c. Electron volt      d. None
The Correct Answer is option b. Joule

Explanation:
 Joule is the SI unit of heat

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MCQs No.52  (Numerical)

The temperature on Celsius scale is 25 degree Celsius. What is the corresponding temperature on the Fahrenheit scale?
a. 40            b. 77 F            c. 50 F            d. 45 F
The Correct Answer is option b. 77 F

Explanation:
By using
F = 9/5(C) + 32.F =77 F

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MCQs No.53 (Numerical)
When 20J of work is done on a gas, 40J of heat energy was released. If the initial internal energy of the gas was 70J. what is the final energy:
a. 50 J          b. 60 J          c. 90 J          d. 110 J
The Correct Answer is option c. 90 J

Explanation:
First law of thermodynamics
∆Q=∆U+∆W
⇒∆U=∆Q-∆W            ∴ ∆U=Uf -Ui

⇒Uf -Ui=∆W-∆Q

or 
Uf =Ui+∆Q-∆W
Uf =70J+40J-20J
Uf =90J

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MCQs No.54 (Numerical)

The temperature of the body is 90°C. its temperature in Fahrenheit is given by:

a. 194°F        b. 164°F        c. 124°F        d. 19.4°F

The Correct Answer is option a. 194°F

Explanation:
Using F = 9/5C + 32.

F=194°F

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MCQs No.55

The equation of state of n moles of an ideal gas is PV= n R T where R is a constant. The SI unit for R is:
a.  JKg⁻¹K⁻¹                  b. Jg⁻¹K⁻¹         

c. JK⁻¹ mol⁻¹                d. JK⁻¹ per molecule 

The Correct Answer is option c. JK⁻¹ mol⁻¹ 

Explanation:
An ideal gas equation

PV = nRT
or
R= PV/nT
 = J/mol·K

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MCQs No.56
In isothermal expansion, the pressure is determined by.

a. Compressibility only    b. temperature only
c. Both a & b                   d. None
The Correct Answer is option 1. Compressibility only"

Explanation:
An ideal gas equation, PV=nRT, in isothermal temperature T is constant. 

so, PV=constant
For such a case, pressure = 1/Compressibility.

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MCQs No.57
A Carnot heat engine absorbs a quantity of heat 1000J from source and does useful work of 300J. The heat rejected is:
a.  300J        b. 1000J        c. 700J        d. 70J
The Correct Answer is option c. 700J
Explanation:

Q2=Q1-W = 1000-300= 700J

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MCQs No.58 (Numerical)
Highest efficiency of heat engine whose lower temperature is 280K and highest temp is 473K is:

a. 70%        b. 100%        c. 35%        d. 40%
The Correct Answer is option d. 40%"

Explanation:
η=1-(T2/T1) x 100
η=1-(280/473) x 100
η= 40%

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MCQs No.59
An inflated tire suddenly bursts, as a result of this temperature of air:
a. increases                        b. Decreases
c. Remains constant            c. May increase or decrease
The Correct Answer is option b. Decreases

Explanation:
The work is done by the gas and the temperature decreases.

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MCQs No.60
Starting with the same initial condition, an ideal gas expands from volume V1 to V2 in three different ways. The work done by the gas is W1 (isothermally), W2 (isobaric) and W3 (adiabatic)
a. W1​>W2​,W3​                b.W1​<W2​=W3​
c. W1>​W2​, W3               c. W2​<W1​=W3​
The Correct Answer is option d. W2​<W1​=W3​"

Explanation:
Curve 1 - isobaric process
Curve 2 - isothermal process
Curve 3 - adiabatic process 

Since work done Area under PV graph
W2​<W1​=W3​

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MCQs No.61 (Numerical)
110J of heat is added gaseous system whose internal energy is 40J. Then the amount of external work done is:

a. 150J        b. 40J        c. 110J        d. 70J
The Correct Answer is option d. 70J"

Explanation:
According to the first law of thermodynamics we have,
ΔQ=∆U+∆W

or 

ΔW=∆Q+∆U

ΔW=110J-40J=70J

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MCQs No.62
When a gas is compressed at constant pressure (isobaric process):
a. Work done on gas is positive
b. Heat is removed from gas
c. Internal energy decrease
d. None
The Correct Answer is option b. Heat is removed from gas

Explanation:

When heat is given to a system at constant pressure it gets expands while in reverse when we remove the heat stove so that its isobarically compressed.

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MCQs No.63 (Numerical)
A Carnot engine takes 300cal of heat at 500K and rejects 150cal of heat to the sink. The temperature of the sink is
a. 100K            b. 700K        c. 125K        d. 250K
The Correct Answer is option d. 250K

Explanation:
Q1/Q2=T1/T2
300/150=500/T2
or 

T2= 250K

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MCQs No.64 (Numerical)

An ideal engine exhausting heat at C is to have 30% efficiency. It must take heat at:
a. 110°C        b. 227°C         c. 327°C        d. 673°C
The Correct Answer is option b. 227°C

Explanation:
η =1-(T2/T1)
T1= 227 C

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MCQs No.65
When a gas is compressed at constant pressure (isobaric process):
a. Work done on gas is positive
b. Heat is removed from gas
c. Internal energy decrease
d. None
The Correct Answer is option b. Heat is removed from gas

Explanation:

When heat is given to a system at constant pressure it gets expands while in reverse when we remove the heat stove so that its isobarically compressed.

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MCQs No.66

Which of the following is NOT a state variable?

a. Temperature                b. Work done
c. Internal energy            d. Entropy
The Correct Answer is option b. Work done

Explanation:

There are 5 state variables i.e. P, V, Entropy, temperature & internal energy

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MCQs No.67 (Numerical)

A thermodynamical system undergoes a process in which its internal energy decreases by 300J. If at same time 120J work is done on the system, then the heat lost by system will be:

a. -420J           b. -170J        c. -360J        d. -120J

The Correct Answer is option a. -420J

Explanation:
Given ∆U= -300J & ∆W= -120J

1st law of thermodynamics 

∆Q=∆U+∆W
∆Q=-300J+-120J
∆Q= -420J

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MCQs No.68

The difference between Cp and Cv is equal to:
a. General Gas constant         b. Planks constant
c. Molar gas constant            c.None
The Correct Answer is option a. General Gas constant

Explanation:
Cp –Cv = R 

where R is general gas constant

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MCQs No.69

Which of the following is true for a closed system?
a. mass entering                 

b. mass leaving
c. mass does not enter or leave the system
d. mass entering can be more or less than the mass leaving
none of the above
The Correct Answer is option c. mass does not enter or leave the system

Explanation:
For a closed system mass does not change.

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MCQs No.70

Work input is directly proportional to heat, and the constant of proportionality is called:

a. joule’s equivalent                b. mechanical equivalent of heat

c. kelvin equivalent                 d. both a & b
The Correct Answer is option d. both a & b

Explanation:

As W= J.Q
or
J=W /Q where J is joule’s equivalent
or mechanical equivalent of heat.

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MCQs No.71

The work done by a closed system in a reversible process is always ___ that done in an irreversible process.
a. less than or more than            b. equal to
c. less than                                d. more than
The Correct Answer is d. more than

Explanation:
A reversible process always produces maximum work.

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MCQs No.72

When a body A is in thermal equilibrium with a body B, and also separately with a body C, then B and C will be in thermal equilibrium with each other.

a. True            b. False            c. Never            d. None

The Correct Answer is option a. True

Explanation:
Explanation: Zeroth law of thermodynamics.

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MCQs No.73 (Numerical)

If a Carnot engine is working with source temperature equal to 227°C and its sink temperature is at 87°C, its efficiency will be:
a. 10%        b. 20%           c. 50%        d. 28%
The Correct Answer is option d. 28%

Explanation:
Given T2=360k T1=500k
For a Carnot engine, efficiency η = 1- (T2/T1) × 100

η =28%

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MCQs No.74 (Numerical)

A Carnot engine, with its cold body at 17 °C has 50% efficiency. If the temperature of its hot body is now increased by 145°C, the efficiency becomes:

a. 55%        b. 60%        c. 40%        d. 45%.

The Correct Answer is option b. 60%

Explanation:
1st case: when T2=17°C =290K, η =50% =1/2, T1=? 

η =1-(T2/T1) 

1/2=1-(290/T1)
or T1=580K

After 145°C increased
T1=725 K, T2=290K
Using η =(1-(T2/T1))x100
η = 60%

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MCQs No.75

First law of thermodynamics corresponds to:
a. Conservation of energy              b. Conservation of mass
c. Conservation of momentum        c. None of these
The Correct Answer is option a. Conservation of energy

Explanation:
Law of conservation of energy.

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MCQs No.76 (Numerical)

The heat reservoir of an ideal Carnot engine is at 800K and its sink is at 400K. The amount of heat taken in it in one second to produce useful mechanical work at the rate of 750 W is:

a. 2250J        b. 1125 J        c. 1500 J        d. 750 J.

The Correct Answer is option c. 1500 J

Explanation:
by using formula of Efficiency = W/Q = 1-(T2/T1)
Q = 1500 joule

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MCQs No.77 (Numerical)

A thermodynamical system undergoes a process in which its internal energy decreases by 200J. If at same time 220J work is done on the system, then the heat lost by system will be:

a. -420J           b. -170J        c. -360J        d. -120J

The Correct Answer is option a. -420J

Explanation:
Given ∆U= -200J & ∆W= -220J

1st law of thermodynamics 

∆Q=∆U+∆W
∆Q=-200J+-220J
∆Q= -420J

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MCQs No.78

In an adiabatic process:
a. ΔQ=ΔU          b. ΔQ=ΔW

c. ΔQ=0            ΔQ=ΔU-ΔW
The Correct Answer is c. ΔQ=0

Explanation:
By definition of adiabatic process ΔQ=0

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MCQs No.79

For a gas obeying Boyle's law if the pressure is doubled, the volume becomes:
a. Double        b. One half        c. Three-fold        d. Remains the same
The Correct Answer is option b. One half

Explanation:
According to Boyle’s law at constant temperature
P α 1/V

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MCQs No.80
The value of the triple point of water is:
a. 273.16°C        b. 273.16°F        c. 273.16 K        d. All of these

The Correct Answer is option c. 273.16 K

Explanation:
Triple point of water=273.16K.

T2= 250K

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MCQs No.81

No entropy change takes place in:
a. Isothermal process                b. Adiabatic process
c. Isobaric process                    c. Isochoric process
The Correct Answer is option b. Adiabatic process

Explanation:
In Adiabatic process ∆Q = 0 mean no change in heat so change in entropy = 0

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MCQs No.82 (Numerical)

A system does 600 J of work and at the same time its internal energy increased by 320J. How much heat has been supplied?
a. 280J        b. 920J        c. 600J        d. 20J
The Correct Answer is option b. 920J

Explanation:
1st law of thermodynamics
∆Q= ∆U+∆W

∆Q= 600J+320J

Q = 920 joule

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MCQs No.83

For Adiabatic process which is true?
a. ΔT=0        b. ΔQ=0        c. ΔV=0            d. none of these
The Correct Answer is option a. ΔT=0

Explanation:
In Adiabatic process the both heat entering and leaving the system is zero so ΔQ=0

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MCQs No.84

Entropy of universe during any natural process:
a. Increases                    b. Increases or remains constant
c. Decreases                   c. Decreases or remains constant
The Correct Answer is option a. Increases

Explanation:
In all natural processes the entropy of the universe is constantly increasing.

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MCQs No.85

Net change in entropy of a system in a Carnot's cycle is:
a. Positive        b. Negative        c. Maximum        d. Zero
The Correct Answer is option d. Zero

Explanation:
In a Carnot engine the net entropy change in a cycle is zero.

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MCQs No.86

If temperature of the heat source is increased, the efficiency of a Carnot engine
a. Increase                            b. decrease        

c. Remain constant                c. First increase than decrease
The Correct Answer is option b. decrease

Explanation:
The efficiency of a Carnot engine increases by increasing the temperature of heat source according to ɳ =1-(T2/T1)

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MCQs No.87

For a closed system difference between heat added and work done by the system
a. ΔV        b. ΔU        c. ΔS        d. ΔP
The Correct Answer is option b. ΔU

Explanation:

from 1st law of thermodynamics equation 

ΔQ=ΔW+ΔU

or

ΔQ−ΔW=ΔU

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MCQs No.88

First law of thermodynamics was presented by
a. Newton        b. Clausius        c. Carnot        d. Pascal
The Correct Answer is option b. Clausius

Explanation:
The first explicit statement of the first law of thermodynamics, by Rudolf Clausius in 1850, referred to cyclic thermodynamic processes.

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MCQs No.89

Total amount of energy in the universe is:
a. increasing                    b. decreasing
c. constant                       c. none of these
The Correct Answer is option c. constant

Explanation:

The First Law of Thermodynamics states that energy can neither be created nor destroyed; it can only be transformed from one form into another. The universe consists of a system and its surroundings taken together. Energy may be transferred between the system and the surroundings in various forms, but it can never be created or destroyed. Therefore, the total amount of energy in the universe remains constant. Consequently, it is impossible to construct a device that can deliver mechanical work without consuming energy.

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MCQs No.90

Entropy remains constant in the process of

 a. Isochoric         b. Isobaric         c. Isothermal         d. Adiabatic

The Correct Answer is option d. Adiabatic

Explanation:

In Adiabatic process ∆Q = 0 mean no change in heat so change in entropy = 0

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MCQs No.91

The value of the adiabatic constant for mono-atomic gas is;

 a. 1.40

 b. 1.44

 c. 1.60

 d. 1.66

The Correct Answer is option d. 1.66

Explanation:

For a monoatomic gas:

$$C_V = \frac{3}{2}R, \quad C_P = \frac{5}{2}R$$

The adiabatic index (ratio of specific heats) is:

$$\gamma = \frac{C_P}{C_V} = \frac{5R/2}{3R/2} = \frac{5}{3} \approx 1.667 \approx 1.66$$

Exam Tip:

For monoatomic gases: $\gamma =5\mathrm{/}3\approx 1.66$

For diatomic gases: $\gamma =7\mathrm{/}5\approx 1.40$

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MCQs No.92

If the temperature of the heat source is increased, the efficiency of Carnot’s engine…
a. Increases                         b. Decreases
c. Remains Constants           d. First increases then become constants.
The Correct Answer is option d. First increases then become constants.

Explanation:
The efficiency of a Carnot engine increases by increasing the temperature of heat source according to ɳ =1-(T2 /T1) x 100

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MCQs No.93

A process in which all the heat energy is used for increasing the internal energy of the system is known as:

 a. Isobaric                    b. Isochoric

 c. Isothermal                 d. Adiabatic

The Correct Answer is option b. Isochoric

Explanation:

The process in which volume remains constant. ΔW= 0 so ΔQ = ΔU

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MCQs No.94

Maximum work can be obtained in the process called______________.
a. Cyclic            b. Isothermal        c. adiabatic        d. Isochoric
The Correct Answer is option a. Cyclic 

Explanation:
In isothermal process the work done is maximum because the total heat is converted in to total work done.

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MCQs No.95

If temperature of the heat source is increased, the efficiency of a Carnot engine______________.
a. Increase                        b. decrease
c. Remain constant            d. first increase than decrease
The Correct Answer is option b. decrease

Explanation:
The efficiency of a Carnot engine increases by increasing the temperature of heat source according to ɳ =1-(T2 /T1) x 100

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MCQs No.96

Maximum work can be obtained in the process called____________.
a. Isothermal              b. adiabatic
c. Isochoric                c. Isobaric
The Correct Answer is option a. Isothermal

Explanation:
In isothermal process temperature is constant so internal energy is zero the total heat converts in to total work done

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MCQs No.97

If the temperature of the heat source is increased, the efficiency of Carnot’s engine_____________.
a. Increases                          b. Decreases
c. Remains Constants            c. First increases then become constants.
The Correct Answer is option c. First increases then become constants.

Explanation:

The efficiency of a Carnot engine increases by increasing the temperature of heat source according to ɳ =1-(T2/T1)

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MCQs No.98

Entropy of a system in a reversible process; 

 a. Decreases

 b. Increases

 c. Infinite

 d. Zero

The Correct Answer is option d. Zero

Explanation:
In a reversible process, entropy change depends only on the initial and final states of the system and not on the path followed. So, in a reversible process, the total entropy change is zero (no net increase in entropy).

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MCQs No.99

In isothermal change, internal energy_____________
a. Increases                            b. Decreases
c. Remains constant                d. Becomes zero.
The Correct Answer is option d. Becomes zero.

Explanation:

Note that the indicator of change in internal energy is change in temperature and in isothermal process temperature remains constant so the internal energy also remains constant.

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MCQs No.100

The efficiency of a heat engine will be 100% when

 a. Engine takes a huge amount of heat from the source

 b. Engine exhaust a very small amount of heat from the sink

 c. The temperature of the cold reservoir is 0°C

 d. The temperature of the cold reservoir is 0K

The Correct Answer is option d. The temperature of the cold reservoir is 0K

Explanation:

For 100% efficiency either the HTR should be at a very much higher temperature or the LTR must be at very lower temperature

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