Home Subjects Mechanical Engineering

Mechanical Engineering

Thermodynamics, hydraulics, machine design

82 Q 3 Topics Take Test
Advertisement
Difficulty: All Easy Medium Hard 71–80 of 82
Topics in Mechanical Engineering
All Thermodynamics 100 Fluid Mechanics 79 Machine Design 80
Q.71 Easy Thermodynamics
Which of the following has the highest specific heat capacity among common substances at room temperature?
A Water
B Iron
C Air
D Copper
Correct Answer:  A. Water
EXPLANATION

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.

Test
Q.72 Easy Thermodynamics
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?
A 300 J
B 700 J
C 250 J
D -300 J
Correct Answer:  A. 300 J
EXPLANATION

From first law: ΔU = Q - W = 500 - 200 = 300 J. Internal energy increases by 300 J.

Test
Q.73 Easy Thermodynamics
Which thermodynamic property is NOT a state function?
A Internal energy
B Enthalpy
C Heat
D Entropy
Correct Answer:  C. Heat
EXPLANATION

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.

Test
Q.74 Easy Thermodynamics
A reversible adiabatic process for an ideal gas is also known as:
A Isothermal process
B Isobaric process
C Isentropic process
D Isochoric process
Correct Answer:  C. Isentropic process
EXPLANATION

A reversible adiabatic process has constant entropy (dS = 0), making it isentropic. This is a key assumption in many thermodynamic analysis for ideal processes.

Test
Q.75 Easy Thermodynamics
What is the relationship between specific heats C_p and C_v for an ideal gas?
A C_p = C_v
B C_p - C_v = R
C C_p + C_v = R
D C_p × C_v = R
Correct Answer:  B. C_p - C_v = R
EXPLANATION

The Mayer relation: C_p - C_v = R, where R is the specific gas constant. This relation holds for all ideal gases.

Test
Q.76 Easy Thermodynamics
What is the first law of thermodynamics in differential form?
A dU = δQ - δW
B dU = δQ + δW
C dU = δQ × δW
D δQ = dU + δW
Correct Answer:  A. dU = δQ - δW
EXPLANATION

The correct form is dU = δQ - δW, where δW = PdV for expansion work. This represents energy conservation in thermodynamic systems.

Test
Q.77 Easy Thermodynamics
A heat engine operates between temperature reservoirs of 500 K and 300 K. What is the maximum possible efficiency of this engine?
A 40%
B 60%
C 50%
D 70%
Correct Answer:  A. 40%
EXPLANATION

Maximum efficiency is Carnot efficiency = 1 - (T_cold/T_hot) = 1 - (300/500) = 0.4 = 40%

Test
Q.78 Easy Thermodynamics
Which process requires zero work to be done by the system?
A Isobaric process
B Isochoric process
C Isothermal process
D Adiabatic process
Correct Answer:  B. Isochoric process
EXPLANATION

In an isochoric (constant volume) process, W = ∫PdV = 0 because volume does not change.

Test
Q.79 Easy Thermodynamics
The entropy of a reversible adiabatic process is:
A Increases
B Decreases
C Remains constant
D Becomes negative
Correct Answer:  C. Remains constant
EXPLANATION

A reversible adiabatic process is also called isentropic, meaning entropy remains constant (ΔS = 0).

Test
Q.80 Easy Thermodynamics
For an ideal gas undergoing an isothermal process, the change in internal energy is:
A Maximum
B Zero
C Negative
D Positive
Correct Answer:  B. Zero
EXPLANATION

In an isothermal process, temperature is constant. For an ideal gas, internal energy depends only on temperature, so ΔU = 0.

Test
IGET
iget AI
Online · Ask anything about exams
Hi! 👋 I'm your iget AI assistant.

Ask me anything about exam prep, MCQ solutions, study tips, or strategies! 🎯
UPSC strategy SSC CGL syllabus Improve aptitude NEET Biology tips