What is the first law of thermodynamics in differential form?
Answer: A
The correct form is dU = δQ - δW, where δW = PdV for expansion work. This represents energy conservation in thermodynamic systems.
Q.24Medium
A piston-cylinder device contains 0.5 kg of steam at 200°C. Heat is removed and the steam condenses to saturated liquid at the same temperature. The latent heat of vaporization at 200°C is 1941 kJ/kg. Calculate heat removed.
Answer: A
Q = m × L_fg = 0.5 kg × 1941 kJ/kg = 970.5 kJ heat is removed during condensation
Q.25Medium
In the Rankine cycle, which process involves expansion in a turbine?
Answer: C
In an ideal Rankine cycle, turbine expansion is isentropic (reversible and adiabatic), maximizing work output. Real turbines follow this closely but with some irreversibilities.
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Q.26Easy
What is the relationship between specific heats C_p and C_v for an ideal gas?
Answer: B
The Mayer relation: C_p - C_v = R, where R is the specific gas constant. This relation holds for all ideal gases.
Q.27Easy
A reversible adiabatic process for an ideal gas is also known as:
Answer: C
A reversible adiabatic process has constant entropy (dS = 0), making it isentropic. This is a key assumption in many thermodynamic analysis for ideal processes.
Q.28Hard
In a gas turbine Brayton cycle, air enters the compressor at 300 K and 100 kPa. The pressure ratio is 8. If γ = 1.4 and R = 287 J/kg·K, find the compressor outlet temperature (assuming isentropic compression).
Which of the following processes is impossible according to the second law of thermodynamics?
Answer: D
The second law states that entropy of an isolated system must increase or remain constant (reversible). A decrease in total entropy violates the second law and is impossible.
Q.30Medium
What is the dryness fraction (quality) of steam at a state where internal energy u = 2400 kJ/kg, u_f = 1317.3 kJ/kg, and u_fg = 1753.7 kJ/kg?
Answer: A
u = u_f + x × u_fg, so x = (u - u_f)/u_fg = (2400 - 1317.3)/1753.7 = 1082.17537.7 ≈ 0.62
Q.31Medium
In a polytropic process PV^n = constant, if n = 1, the process is:
Answer: B
For polytropic process with n=1: PV = constant, which is the ideal gas law at constant temperature, making it isothermal (T = constant).
Q.32Hard
A Carnot heat pump operates between 270 K and 330 K. If 1000 J of work is supplied, how much heat is delivered to the hot reservoir?
Answer: C
COP_heating = T_H/(T_H - T_C) = 330/(330-270) = 60330 = 5.5. Q_H = W × COP = 1000 × 5.5 = 5500 J. Adding work input: Total = 5500 + 500 = 6000 J
Q.33Easy
Which thermodynamic property is NOT a state function?
Answer: C
Heat (Q) and work (W) are path-dependent quantities, not state functions. Internal energy, enthalpy, and entropy are state functions depending only on initial and final states.
Q.34Medium
In an Otto cycle, the compression ratio is 10 and γ = 1.4. Calculate the thermal efficiency.
A gas undergoes an isobaric process. If 500 J of heat is added and the gas expands such that work done by the gas is 200 J, what is the change in internal energy?
Answer: A
From first law: ΔU = Q - W = 500 - 200 = 300 J. Internal energy increases by 300 J.
Q.36Medium
At what condition is the specific heat capacity at constant pressure equal to infinity?
Answer: B
During phase transitions (like vaporization at constant T and P), infinite heat can be absorbed without temperature change, making C_p → ∞.
Q.37Easy
Which of the following has the highest specific heat capacity among common substances at room temperature?
Answer: A
Water has exceptionally high specific heat (~4.18 kJ/kg·K), much higher than metals (iron ~0.46) and air (~1.01). This is due to hydrogen bonding in water.
Q.38Easy
In a constant volume process, 300 J of heat is removed from a gas. Calculate the work done and change in internal energy.
Answer: A
In constant volume process: W = ∫PdV = 0. From first law: ΔU = Q - W = -300 - 0 = -300 J
Q.39Medium
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?
Answer: A
Efficiency η = W/Q_in = (Q_in - Q_out)/Q_in = (2000-1200)/2000 = 2000800 = 0.4 = 40%. Work output W = 800 J
Q.40Medium
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?
Answer: A
Maximum efficiency occurs in a Carnot engine: η = 1 - (T_cold/T_hot) = 1 - (500300) = 1 - 0.6 = 0.4 or 40%
