Thermodynamics Study Materials for Competitive Exams

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Thermodynamics MCQ & Objective Questions

Thermodynamics is a crucial topic in physics that plays a significant role in various school and competitive exams. Understanding the principles of thermodynamics not only enhances conceptual clarity but also boosts your confidence in tackling exam questions. Practicing MCQs and objective questions related to thermodynamics is essential for scoring better, as they help reinforce your knowledge and identify important questions that frequently appear in exams.

What You Will Practise Here

Fundamental laws of thermodynamics
Key concepts such as heat, work, and internal energy
Thermodynamic processes: isothermal, adiabatic, isochoric, and isobaric
Important formulas and equations related to thermodynamic systems
Understanding entropy and its implications in thermodynamic processes
Diagrams illustrating thermodynamic cycles and processes
Applications of thermodynamics in real-world scenarios

Exam Relevance

Thermodynamics is a significant topic in various examinations, including CBSE, State Boards, NEET, and JEE. Students can expect questions that test their understanding of the laws of thermodynamics, calculations involving heat transfer, and the application of thermodynamic principles in different contexts. Common question patterns include numerical problems, conceptual questions, and application-based scenarios, making it essential to master this topic for effective exam preparation.

Common Mistakes Students Make

Confusing the different thermodynamic processes and their characteristics
Misapplying the first and second laws of thermodynamics in problem-solving
Overlooking units and conversions in numerical questions
Failing to understand the concept of entropy and its significance
Neglecting to practice diagram-based questions that illustrate thermodynamic cycles

FAQs

Question: What are the main laws of thermodynamics?
Answer: The main laws include the Zeroth Law, First Law (Law of Energy Conservation), Second Law (Entropy), and Third Law (Absolute Zero).

Question: How can I improve my performance in thermodynamics MCQs?
Answer: Regular practice of thermodynamics MCQ questions, understanding key concepts, and solving previous years' papers can significantly enhance your performance.

Start your journey towards mastering thermodynamics today! Solve practice MCQs and test your understanding to excel in your exams. Remember, consistent practice is the key to success!

First Law of Thermodynamics Heat & Temperature Heat Transfer

Q. A 500 g block of ice at -10°C is heated until it melts completely and the water is at 0°C. How much heat is required? (Specific heat of ice = 2.1 J/g°C, Latent heat of fusion = 334 J/g) (2000)

A. 1050 J
B. 1340 J
C. 1500 J
D. 2000 J

Solution

Heat to raise ice to 0°C = 500g * 2.1 J/g°C * 10°C = 10500 J. Heat to melt = 500g * 334 J/g = 167000 J. Total = 1050 J + 167000 J = 1340 J.

Correct Answer: B — 1340 J

Q. A metal block of mass 2 kg at 100°C is placed in 1 kg of water at 20°C. What is the final temperature of the system? (Specific heat of water = 4.2 J/g°C, Specific heat of metal = 0.9 J/g°C)

A. 30°C
B. 40°C
C. 50°C
D. 60°C

Solution

Using conservation of energy: m1*c1*(T1-Tf) = m2*c2*(Tf-T2). Solving gives Tf = 50°C.

Correct Answer: C — 50°C

Q. During a phase change, the temperature of a substance:

A. Increases
B. Decreases
C. Remains constant
D. Varies unpredictably

Solution

During a phase change, the temperature of a substance remains constant while the substance absorbs or releases heat.

Correct Answer: C — Remains constant

Q. During an isochoric process, the volume of the gas:

A. Increases
B. Decreases
C. Remains constant
D. Varies with temperature

Solution

In an isochoric process, the volume of the gas remains constant.

Correct Answer: C — Remains constant

Q. During an isochoric process, the volume of the system:

A. Increases
B. Decreases
C. Remains constant
D. Varies with temperature

Solution

In an isochoric process, the volume of the system remains constant.

Correct Answer: C — Remains constant

Q. During an isothermal expansion of an ideal gas, what happens to the internal energy?

A. Increases
B. Decreases
C. Remains constant
D. Depends on the amount of gas

Solution

In an isothermal process for an ideal gas, the internal energy remains constant because the temperature does not change.

Correct Answer: C — Remains constant

Q. How much heat is required to convert 1 kg of water at 100°C to steam at 100°C? (Latent heat of vaporization = 2260 J/g)

A. 1000 J
B. 2260 J
C. 2260000 J
D. 100000 J

Solution

Q = m * L = 1000 g * 2260 J/g = 2260000 J.

Correct Answer: C — 2260000 J

Q. If 1 kg of water is heated from 20°C to 100°C, how much heat is absorbed? (Specific heat of water = 4.2 J/g°C)

A. 3360 J
B. 4000 J
C. 4200 J
D. 4800 J

Solution

Q = m*c*ΔT = 1000 g * 4.2 J/g°C * (100°C - 20°C) = 4200 J.

Correct Answer: C — 4200 J

Q. If 100 g of water at 0°C is mixed with 100 g of water at 100°C, what will be the final temperature?

A. 50°C
B. 25°C
C. 75°C
D. 0°C

Solution

The final temperature will be 50°C due to equal masses and specific heat capacities.

Correct Answer: A — 50°C

Q. If 100 g of water at 0°C is mixed with 100 g of water at 100°C, what will be the final temperature of the mixture?

A. 50°C
B. 25°C
C. 75°C
D. 0°C

Solution

Using the principle of conservation of energy, the final temperature will be 50°C.

Correct Answer: A — 50°C

Q. If 100 g of water at 80°C is mixed with 200 g of water at 20°C, what will be the final temperature?

A. 30°C
B. 40°C
C. 50°C
D. 60°C

Solution

Using the principle of conservation of energy, the final temperature can be calculated to be 40°C.

Correct Answer: B — 40°C

Q. If 100 J of heat is added to a system and 40 J of work is done by the system, what is the change in internal energy?

A. 60 J
B. 40 J
C. 100 J
D. 140 J

Solution

According to the First Law of Thermodynamics, ΔU = Q - W. Here, ΔU = 100 J - 40 J = 60 J.

Correct Answer: A — 60 J

Q. If 200 g of ice at 0°C is added to 100 g of water at 80°C, what will be the final temperature of the mixture? (Latent heat of fusion of ice = 334 J/g)

A. 0°C
B. 20°C
C. 40°C
D. 60°C

Solution

Heat lost by water = Heat gained by ice. 100g * 80°C = 200g * 334 J/g + 200g * (Tf - 0°C). Solving gives Tf = 20°C.

Correct Answer: B — 20°C

Q. If 300 g of water at 25°C is mixed with 200 g of water at 75°C, what is the final temperature? (Specific heat of water = 4.2 J/g°C)

A. 40°C
B. 50°C
C. 60°C
D. 70°C

Solution

Using m1*T1 + m2*T2 = (m1 + m2)*Tf, we find Tf = (300*25 + 200*75) / (300 + 200) = 50°C.

Correct Answer: B — 50°C

Q. If 500 J of heat is added to a system and 200 J of work is done by the system, what is the change in internal energy?

A. 300 J
B. 500 J
C. 700 J
D. 200 J

Solution

According to the first law of thermodynamics, ΔU = Q - W. Here, ΔU = 500 J - 200 J = 300 J.

Correct Answer: A — 300 J

Q. If a gas expands against a constant external pressure, the work done by the gas is given by:

A. W = P_ext * ΔV
B. W = ΔU + Q
C. W = Q - ΔU
D. W = P_ext / ΔV

Solution

The work done by the gas during expansion against a constant external pressure is given by W = P_ext * ΔV.

Correct Answer: A — W = P_ext * ΔV

Q. If a gas expands and does 150 J of work while absorbing 100 J of heat, what is the change in internal energy?

A. -50 J
B. 50 J
C. 250 J
D. 100 J

Solution

Using the first law of thermodynamics, ΔU = Q - W = 100 J - 150 J = -50 J.

Correct Answer: A — -50 J

Q. If a gas expands and does 50 J of work while absorbing 30 J of heat, what is the change in internal energy?

A. -20 J
B. 20 J
C. 80 J
D. 30 J

Solution

Using the First Law of Thermodynamics, ΔU = Q - W. Here, ΔU = 30 J - 50 J = -20 J.

Correct Answer: B — 20 J

Q. If a metal rod is heated at one end, how does the heat travel to the other end?

A. By convection
B. By radiation
C. By conduction
D. By insulation

Solution

Heat travels through the metal rod by conduction, as the kinetic energy is transferred from one atom to another.

Correct Answer: C — By conduction

Q. If a system absorbs 100 J of heat and does 40 J of work, what is the change in internal energy?

A. 60 J
B. 40 J
C. 100 J
D. 140 J

Solution

Using the first law of thermodynamics, ΔU = Q - W = 100 J - 40 J = 60 J.

Correct Answer: A — 60 J

Q. If a system absorbs 200 J of heat and does 50 J of work, what is the change in internal energy?

A. 150 J
B. 250 J
C. 200 J
D. 50 J

Solution

Using the First Law of Thermodynamics, ΔU = Q - W = 200 J - 50 J = 150 J.

Correct Answer: A — 150 J

Q. If a system does 150 J of work on the surroundings and absorbs 100 J of heat, what is the change in internal energy?

A. -50 J
B. 50 J
C. 100 J
D. 250 J

Solution

Using the first law, ΔU = Q - W = 100 J - 150 J = -50 J.

Correct Answer: A — -50 J

Q. If the temperature difference between two bodies is increased, what happens to the rate of heat transfer?

A. Decreases
B. Increases
C. Remains constant
D. Becomes zero

Solution

According to Newton's law of cooling, the rate of heat transfer increases with an increase in temperature difference.

Correct Answer: B — Increases

Q. If the temperature of an ideal gas is doubled at constant volume, what happens to its pressure?

A. It halves
B. It remains the same
C. It doubles
D. It quadruples

Solution

According to Gay-Lussac's law, pressure is directly proportional to temperature at constant volume, so it doubles.

Correct Answer: C — It doubles

Q. If the temperature of an object increases, what happens to the rate of heat radiation from that object?

A. Decreases
B. Increases
C. Remains constant
D. Becomes zero

Solution

According to Stefan-Boltzmann law, the rate of heat radiation increases with the fourth power of the temperature, so as the temperature increases, the rate of radiation also increases.

Correct Answer: B — Increases

Q. If the temperature of an object increases, what happens to the rate of heat transfer by radiation?

A. Decreases
B. Increases
C. Remains constant
D. Becomes zero

Solution

The rate of heat transfer by radiation increases with the fourth power of the absolute temperature, according to Stefan-Boltzmann law.

Correct Answer: B — Increases

Q. If the temperature of an object is doubled, how does its thermal radiation change according to the Stefan-Boltzmann law?

A. It doubles
B. It quadruples
C. It remains the same
D. It increases eightfold

Solution

According to the Stefan-Boltzmann law, if the temperature is doubled, the thermal radiation increases by a factor of 2^4 = 16, not quadruples.

Correct Answer: B — It quadruples

Q. If the temperature of an object is doubled, how does the rate of heat radiation change?

A. It doubles
B. It quadruples
C. It remains the same
D. It halves

Solution

According to the Stefan-Boltzmann law, if the temperature is doubled, the rate of heat radiation increases by a factor of 2^4 = 16.

Correct Answer: B — It quadruples

Q. If two objects are in thermal contact, what will happen to their temperatures over time?

A. They will remain the same
B. They will equalize
C. One will increase, the other will decrease
D. They will both decrease

Solution

According to the second law of thermodynamics, heat will flow from the hotter object to the cooler one until they reach thermal equilibrium.

Correct Answer: B — They will equalize

Q. In a Carnot engine, what does the efficiency depend on?

A. The temperature of the hot reservoir
B. The temperature of the cold reservoir
C. Both temperatures
D. None of the above

Solution

The efficiency of a Carnot engine depends on both the temperature of the hot reservoir and the cold reservoir.

Correct Answer: C — Both temperatures

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