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Mechanical Engineering
Thermodynamics

Thermodynamics, hydraulics, machine design

45 Q 3 Topics Take Mock Test
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Difficulty: All Easy Medium Hard 21–30 of 45
Topics in Mechanical Engineering
All Thermodynamics 100 Fluid Mechanics 79 Machine Design 80
Q.21 Medium Thermodynamics
Steam at 80 kPa with enthalpy 2660 kJ/kg and entropy 7.31 kJ/kg·K. What is the thermodynamic state of this steam?
A Saturated liquid
B Saturated vapor
C Wet steam (two-phase mixture)
D Superheated vapor
Correct Answer:  D. Superheated vapor
EXPLANATION

At 80 kPa, saturated vapor properties are h_g ≈ 2660.3 kJ/kg and s_g ≈ 7.31 kJ/kg·K. Given values match saturated vapor line, indicating superheated or at saturation point; properties at saturation boundary indicate superheated vapor.

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Q.22 Medium Thermodynamics
A heat pump operating between 280 K and 350 K has a coefficient of performance (COP) of 5. How much work input is required to transfer 1000 J of heat to the hot reservoir?
A 200 J
B 250 J
C 300 J
D 175 J
Correct Answer:  A. 200 J
EXPLANATION

COP = Q_h / W = 5. Therefore, W = Q_h / COP = 1000 / 5 = 200 J

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Q.23 Medium Thermodynamics
For an ideal gas in a constant pressure (isobaric) process from state 1 to state 2, the change in specific entropy is:
A Cp ln(T2/T1)
B Cv ln(V2/V1)
C R ln(P2/P1)
D Cp ln(P2/P1)
Correct Answer:  A. Cp ln(T2/T1)
EXPLANATION

For isobaric process: dS = dQ/T = nCp dT/T. Integrating: ΔS = nCp ln(T2/T1) or specific entropy Δs = Cp ln(T2/T1)

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Q.24 Medium Thermodynamics
In a turbocharger application, compressed air enters a turbine. If the isentropic efficiency of the turbine is 0.85, what does this indicate?
A 85% of input heat is converted to work
B Actual work output is 85% of isentropic work output
C 85% of the system is reversible
D The process is 85% adiabatic
Correct Answer:  B. Actual work output is 85% of isentropic work output
EXPLANATION

Isentropic efficiency = Actual work output / Isentropic work output = 0.85. It compares real process with ideal isentropic process.

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Q.25 Medium Thermodynamics
During throttling of a real gas through an expansion valve, which thermodynamic property remains constant?
A Temperature
B Entropy
C Enthalpy
D Internal energy
Correct Answer:  C. Enthalpy
EXPLANATION

Throttling is an isenthalpic process (constant enthalpy). Temperature may change for real gases due to Joule-Thomson effect.

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Q.26 Medium Thermodynamics
During an adiabatic expansion of an ideal gas, the temperature of the gas decreases. This is because:
A Heat is removed from the gas
B The gas does work on its surroundings, reducing internal energy
C The surroundings do work on the gas
D The gas absorbs heat from surroundings
Correct Answer:  B. The gas does work on its surroundings, reducing internal energy
EXPLANATION

In adiabatic process, Q = 0. From first law: dU = -W. During expansion, W > 0, so dU < 0, meaning internal energy and temperature decrease.

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Q.27 Medium Thermodynamics
Which of the following statements is correct regarding the first law of thermodynamics?
A Energy can be created in an isolated system
B The internal energy of an isolated system can increase indefinitely
C Heat and work are forms of energy transfer, not properties of a system
D Internal energy is path-dependent
Correct Answer:  C. Heat and work are forms of energy transfer, not properties of a system
EXPLANATION

The first law states that dU = Q - W, where heat (Q) and work (W) are energy transfer mechanisms, not state properties. Internal energy (U) is a state property.

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Q.28 Medium Thermodynamics
A reversible heat engine operates between two thermal reservoirs. If the temperature of the hot reservoir is 500 K and the cold reservoir is 300 K, what is the maximum possible efficiency of this engine?
A 40%
B 60%
C 50%
D 75%
Correct Answer:  A. 40%
EXPLANATION

Maximum efficiency occurs in a Carnot engine: η = 1 - (T_cold/T_hot) = 1 - (300/500) = 1 - 0.6 = 0.4 or 40%

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Q.29 Medium Thermodynamics
A heat engine receives 2000 J from a hot reservoir and rejects 1200 J to a cold reservoir. What is the thermal efficiency and work output?
A η = 40%, W = 800 J
B η = 60%, W = 1200 J
C η = 40%, W = 1200 J
D η = 50%, W = 1000 J
Correct Answer:  A. η = 40%, W = 800 J
EXPLANATION

Efficiency η = W/Q_in = (Q_in - Q_out)/Q_in = (2000-1200)/2000 = 800/2000 = 0.4 = 40%. Work output W = 800 J

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Q.30 Medium Thermodynamics
At what condition is the specific heat capacity at constant pressure equal to infinity?
A At critical point
B During phase change at constant temperature and pressure
C At absolute zero
D At very high pressures
Correct Answer:  B. During phase change at constant temperature and pressure
EXPLANATION

During phase transitions (like vaporization at constant T and P), infinite heat can be absorbed without temperature change, making C_p → ∞.

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