Below is a sample revision test designed around many of the core topics found
in “Thermodynamics and Heat Power, 8th Edition” by Granet. Each question is
followed by an explanation (rationale) so you can review not only the correct
answer but also the reasoning behind it.
Revision Test for Thermodynamics and Heat Power
Question 1: The First Law of Thermodynamics
A closed system undergoes a process in which 400 kJ of heat is added while 150 kJ of work is done by
the system. What is the change in its internal energy?
A. 550 kJ increase
B. 250 kJ increase
C. 250 kJ decrease
D. 550 kJ decrease
Correct Answer: B. 250 kJ increase
Rationale:
The first law states that the change in internal energy is equal to the heat added to the system minus
the work done by the system, i.e., ΔU = Q – W. Here, ΔU = 400 kJ – 150 kJ = 250 kJ. An increase in energy
indicates that the system has stored more energy from the heat input than it used to perform work.
Question 2: Second Law and Efficiency of Heat Engines
Which of the following statements best summarizes the second law of thermodynamics in the context of
heat engines?
A. No process is possible whose sole result is the absorption of heat from a reservoir and the conversion
of all that heat into work.
B. Energy can neither be created nor destroyed.
C. Entropy always decreases in an isolated system.
D. A perpetual motion machine of the second kind is feasible with improved design.
Correct Answer: A. No process is possible whose sole result is the absorption of heat from a reservoir
and the conversion of all that heat into work.
Rationale:
The second law of thermodynamics imposes a fundamental limit on the efficiency of heat engines by
stating that some heat must always be rejected to a sink (or low-temperature reservoir). This means
that 100% conversion of heat into work is impossible, which is exactly what option A states.
Question 3: Carnot Cycle Efficiency
, The efficiency of a Carnot engine is given by η = 1 – (Tₗ/Tₕ), where Tₗ and Tₕ are the absolute
temperatures of the cold and hot reservoirs, respectively. If Tₕ = 600 K and Tₗ = 300 K, what is the
maximum possible efficiency?
A. 25%
B. 50%
C. 75%
D. 100%
Correct Answer: B. 50%
Rationale:
Using the Carnot efficiency formula, η = 1 – (300/600) = 1 – 0.5 = 0.5 or 50%. This represents the
maximum theoretical efficiency any engine operating between these two temperatures can achieve.
Question 4: Properties of an Ideal Gas
For an ideal gas undergoing a constant pressure process, which of the following correctly relates the
work done (W) to the change in volume (ΔV) and the pressure (P)?
A. W = PΔV
B. W = ΔV/P
C. W = P/ΔV
D. W = ΔV²/P
Correct Answer: A. W = PΔV
Rationale:
Under constant pressure, the work done by a gas during expansion (or on the gas during compression) is
defined as W = PΔV. This direct relationship is one of the fundamental results derived from the work
integral in thermodynamics.
Question 5: Isentropic Processes
In an isentropic process for an ideal gas, which of the following remains constant?
A. Temperature
B. Pressure
C. Volume
D. Entropy
Correct Answer: D. Entropy
Rationale:
By definition, an isentropic process is both adiabatic (no heat transfer) and reversible, meaning the
entropy of the system remains constant throughout the process. Temperature, pressure, and volume
can change as long as the process does not generate entropy.
in “Thermodynamics and Heat Power, 8th Edition” by Granet. Each question is
followed by an explanation (rationale) so you can review not only the correct
answer but also the reasoning behind it.
Revision Test for Thermodynamics and Heat Power
Question 1: The First Law of Thermodynamics
A closed system undergoes a process in which 400 kJ of heat is added while 150 kJ of work is done by
the system. What is the change in its internal energy?
A. 550 kJ increase
B. 250 kJ increase
C. 250 kJ decrease
D. 550 kJ decrease
Correct Answer: B. 250 kJ increase
Rationale:
The first law states that the change in internal energy is equal to the heat added to the system minus
the work done by the system, i.e., ΔU = Q – W. Here, ΔU = 400 kJ – 150 kJ = 250 kJ. An increase in energy
indicates that the system has stored more energy from the heat input than it used to perform work.
Question 2: Second Law and Efficiency of Heat Engines
Which of the following statements best summarizes the second law of thermodynamics in the context of
heat engines?
A. No process is possible whose sole result is the absorption of heat from a reservoir and the conversion
of all that heat into work.
B. Energy can neither be created nor destroyed.
C. Entropy always decreases in an isolated system.
D. A perpetual motion machine of the second kind is feasible with improved design.
Correct Answer: A. No process is possible whose sole result is the absorption of heat from a reservoir
and the conversion of all that heat into work.
Rationale:
The second law of thermodynamics imposes a fundamental limit on the efficiency of heat engines by
stating that some heat must always be rejected to a sink (or low-temperature reservoir). This means
that 100% conversion of heat into work is impossible, which is exactly what option A states.
Question 3: Carnot Cycle Efficiency
, The efficiency of a Carnot engine is given by η = 1 – (Tₗ/Tₕ), where Tₗ and Tₕ are the absolute
temperatures of the cold and hot reservoirs, respectively. If Tₕ = 600 K and Tₗ = 300 K, what is the
maximum possible efficiency?
A. 25%
B. 50%
C. 75%
D. 100%
Correct Answer: B. 50%
Rationale:
Using the Carnot efficiency formula, η = 1 – (300/600) = 1 – 0.5 = 0.5 or 50%. This represents the
maximum theoretical efficiency any engine operating between these two temperatures can achieve.
Question 4: Properties of an Ideal Gas
For an ideal gas undergoing a constant pressure process, which of the following correctly relates the
work done (W) to the change in volume (ΔV) and the pressure (P)?
A. W = PΔV
B. W = ΔV/P
C. W = P/ΔV
D. W = ΔV²/P
Correct Answer: A. W = PΔV
Rationale:
Under constant pressure, the work done by a gas during expansion (or on the gas during compression) is
defined as W = PΔV. This direct relationship is one of the fundamental results derived from the work
integral in thermodynamics.
Question 5: Isentropic Processes
In an isentropic process for an ideal gas, which of the following remains constant?
A. Temperature
B. Pressure
C. Volume
D. Entropy
Correct Answer: D. Entropy
Rationale:
By definition, an isentropic process is both adiabatic (no heat transfer) and reversible, meaning the
entropy of the system remains constant throughout the process. Temperature, pressure, and volume
can change as long as the process does not generate entropy.
