Second Law of Thermodynamics - UNSOLVED PRACTICE SET
Chapter: Thermodynamics | Topic: Second Law of Thermodynamics
SECOND LAW OF THERMODYNAMICS - UNSOLVED PRACTICE SET
Topic: Second Law of Thermodynamics
Multiple Choice Questions
Q1. The Second Law of Thermodynamics states that:
- Energy cannot be created or destroyed
- The entropy of the universe increases in a spontaneous process
- The entropy of a perfect crystal is zero at absolute zero
- Heat flows from cold to hot objects spontaneously
Q2. For a reversible process, the entropy change of the universe is:
- Positive
- Negative
- Zero
- Infinite
Q3. Which of the following statements is consistent with the Second Law of Thermodynamics?
- Heat can spontaneously flow from a colder body to a hotter body
- It is possible to convert heat completely into work in a cyclic process
- The total entropy of an isolated system always increases for a spontaneous process
- Perpetual motion machines of the first kind are possible
Q4. The efficiency of a Carnot engine depends on:
- The working substance used
- The temperatures of the hot and cold reservoirs only
- The amount of heat supplied
- The pressure of the working substance
Q5. A process is thermodynamically reversible when:
- It occurs rapidly
- It is carried out infinitesimally slowly, maintaining equilibrium at every stage
- It produces maximum entropy
- It is carried out at constant temperature
Q6. The coefficient of performance of a refrigerator is:
- Always greater than 1
- Always less than 1
- Equal to 1
- Equal to the efficiency of a heat engine
Short Answer Questions
Q7. State the Second Law of Thermodynamics in terms of entropy. Why is it considered a 'law of nature' rather than a 'law of systems'?
Q8. Explain why it is impossible to build a perpetual motion machine of the second kind using the Second Law of Thermodynamics.
Q9. Differentiate between a reversible process and an irreversible process with one example each. Which one is more efficient?
Q10. Why does the Second Law of Thermodynamics imply that all real processes are irreversible? Explain with an example.
Q11. A student claims that a heat engine can convert 100% of the heat supplied into work. Use the Second Law to explain why this is impossible.
Q12. Explain the concept of a heat reservoir. Why is the ocean considered an almost ideal heat reservoir?
Long Answer Questions
Q13. (a) State the Kelvin-Planck and Clausius statements of the Second Law of Thermodynamics. Show that these two statements are equivalent.
(b) Explain why the Second Law of Thermodynamics places a fundamental limit on the efficiency of heat engines.
(c) A student designs a heat engine that absorbs 1000 J of heat and does 900 J of work. Is this design feasible? Explain using the Second Law.
Q14. (a) Define the efficiency of a Carnot engine and derive its expression:
ฮท = 1 - Tc / Th(b) A Carnot engine operates between 600 K and 300 K. Calculate its efficiency.
(c) Can the efficiency of a Carnot engine ever reach 100%? Explain why or why not.
(d) Why is the Carnot engine considered an ideal engine, even though it cannot be practically constructed?
Q15. (a) Explain the working principle of a refrigerator using the Second Law of Thermodynamics.
(b) Define the coefficient of performance (COP) of a refrigerator.
(c) A refrigerator maintains its interior at 277 K while the room temperature is 300 K. Calculate the maximum possible COP.
(d) In India, refrigerators are essential appliances in almost every household. How does understanding the Second Law help in designing more energy-efficient refrigerators?
Numerical / Application-Based Problems
Q16. A heat engine operates between a hot reservoir at 800 K and a cold reservoir at 300 K.
(a) Calculate the maximum possible efficiency of this engine (Carnot efficiency).
(b) If the engine absorbs 2000 J of heat from the hot reservoir, calculate the maximum work output and the heat rejected to the cold reservoir.
(c) A student claims that by using a better working substance, the efficiency can be increased to 70%. Is this claim valid? Explain.
Q17. A Carnot refrigerator operates between 263 K (freezer) and 298 K (room).
(a) Calculate the coefficient of performance (COP) of this refrigerator.
(b) If 5000 J of heat is to be removed from the freezer, calculate the minimum work input required.
(c) Calculate the heat rejected to the room.
(d) In summer, when the room temperature rises to 310 K, how does the COP change? Calculate the new COP and explain why refrigerators consume more electricity in summer.
Q18. In India, thermal power plants generate electricity by using steam at high temperatures. Consider a power plant that uses steam at 800 K and rejects heat to a river at 300 K.
(a) Calculate the maximum theoretical efficiency of this power plant.
(b) If the plant generates 500 MW of power, calculate the minimum rate of heat absorption from the steam and the minimum rate of heat rejection to the river.
(c) The actual efficiency of most Indian thermal power plants is around 30-35%. Explain why the actual efficiency is much lower than the theoretical maximum.
(d) Suggest two ways to improve the efficiency of thermal power plants in India, keeping in mind environmental concerns.