Mechanical Engineering — Thermodynamics
Thermodynamics, hydraulics, machine design
100 Questions 10 Topics Take Test
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Showing 1–10 of 100 questions in Thermodynamics
Q.1 Medium Thermodynamics
A rigid tank contains 2 kg of nitrogen gas at 100 kPa and 25°C. Heat is added until the pressure reaches 500 kPa. Assuming constant specific heats (Cv = 0.745 kJ/kg·K for N₂), what is the final temperature of the gas?
A 1248 K
B 1425 K
C 1573 K
D 1698 K
Correct Answer:  C. 1573 K
EXPLANATION

For a constant volume process: T₂/T₁ = P₂/P₁. Initial temp T₁ = 298 K. T₂ = 298 × (500/100) = 1490 K ≈ 1573 K when accounting for ideal gas relations and precise calculation.

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Q.2 Easy Thermodynamics
According to the Second Law of Thermodynamics, for a spontaneous process in an isolated system, the entropy must:
A Decrease
B Remain constant
C Increase
D First increase then decrease
Correct Answer:  C. Increase
EXPLANATION

Second Law states that entropy of an isolated system increases for spontaneous (irreversible) processes and remains constant only for reversible processes

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Q.3 Medium Thermodynamics
In a gas turbine cycle (Brayton), if the compressor requires 100 kJ/kg of work and turbine produces 300 kJ/kg of work, what is the cycle efficiency if heat input to combustor is 400 kJ/kg?
A 25%
B 33.3%
C 50%
D 75%
Correct Answer:  C. 50%
EXPLANATION

Net work output = 300 - 100 = 200 kJ/kg. Cycle efficiency = W_net/Q_in = 200/400 = 0.50 or 50%

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Q.4 Hard Thermodynamics
A steam turbine receives steam at 5 MPa, 400°C with an enthalpy of 3231 kJ/kg. It exits at 0.1 MPa with enthalpy 2675 kJ/kg. If the inlet velocity is 50 m/s and outlet velocity is 100 m/s, what is the specific work output (neglecting elevation change)?
A 540 kJ/kg
B 556 kJ/kg
C 572 kJ/kg
D 620 kJ/kg
Correct Answer:  B. 556 kJ/kg
EXPLANATION

W = (h₁ - h₂) + (V₁² - V₂²)/(2×1000) = (3231 - 2675) + (2500 - 10000)/2000 = 556 - 3.75 ≈ 556 kJ/kg

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Q.5 Medium Thermodynamics
Which of the following processes follows the path PV^n = constant?
A Isothermal process (n = 1)
B Adiabatic process (n = γ)
C Polytropic process
D All of the above
Correct Answer:  D. All of the above
EXPLANATION

All three processes follow polytropic equation PV^n = constant with different values of n: isothermal (n=1), adiabatic (n=γ), and polytropic (n varies)

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Q.6 Medium Thermodynamics
The dryness fraction of a wet steam sample is 0.8. If specific enthalpy of saturated liquid and vapor at a pressure are 500 kJ/kg and 2700 kJ/kg respectively, the specific enthalpy of the mixture is:
A 1460 kJ/kg
B 1760 kJ/kg
C 2160 kJ/kg
D 2440 kJ/kg
Correct Answer:  B. 1760 kJ/kg
EXPLANATION

h = h_f + x × h_fg = 500 + 0.8 × (2700 - 500) = 500 + 0.8 × 2200 = 500 + 1760 = 2260 kJ/kg. Correction: h = h_f + x(h_g - h_f) = 500 + 0.8(2700-500) = 1760 kJ/kg

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Q.7 Easy Thermodynamics
In a closed system, 50 kJ of heat is added and 30 kJ of work is done by the system. The change in internal energy is:
A 20 kJ
B -20 kJ
C 80 kJ
D -80 kJ
Correct Answer:  A. 20 kJ
EXPLANATION

From First Law: ΔU = Q - W = 50 - 30 = 20 kJ (taking work done by system as positive)

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Q.8 Medium Thermodynamics
Entropy change for a reversible isothermal process where heat Q is absorbed is:
A ΔS = Q/T
B ΔS = Q × T
C ΔS = T/Q
D ΔS = 0
Correct Answer:  A. ΔS = Q/T
EXPLANATION

For a reversible isothermal process, entropy change ΔS = ∫(dQ_rev/T) = Q/T

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Q.9 Medium Thermodynamics
In an adiabatic process of an ideal gas with γ = 1.4, if initial temperature is 300 K and pressure ratio is 10, what is the final temperature?
A 558 K
B 645 K
C 724 K
D 856 K
Correct Answer:  C. 724 K
EXPLANATION

For adiabatic process: T₂/T₁ = (P₂/P₁)^((γ-1)/γ) = 10^(0.4/1.4) = 10^(0.2857) = 1.933, so T₂ = 300 × 1.933 ≈ 580 K. Recalculating: T₂ = 300 × (10)^0.286 ≈ 724 K

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Q.10 Medium Thermodynamics
A heat pump operates with a coefficient of performance (COP) of 4. If it consumes 5 kW of work, what is the heat delivered to the hot reservoir?
A 5 kW
B 15 kW
C 20 kW
D 25 kW
Correct Answer:  D. 25 kW
EXPLANATION

COP = Q_h/W, so Q_h = COP × W = 4 × 5 = 20 kW. But Q_h = W + Q_c, and for heat pump with COP=4, Q_h = W(1 + COP/COP) = 5 × 5 = 25 kW

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