By Fatskills Exam Guides Team — the exam nerds behind 28,500+ quizzes and 2.1M practice questions across 500+ global exams.
The First and Second Laws of Thermodynamics are fundamental concepts that govern the behavior of energy and its interactions. These laws appear in 2-3 questions every year, making them a crucial part of your JEE preparation. The difficulty level is moderate, with a slight bias towards JEE Advanced.
Question 1: A refrigerator operates between 0°C and 25°C. The coefficient of performance (COP) is defined as the ratio of heat removed to work done. What is the COP of the refrigerator?
A) 2.5 B) 5 C) 10 D) 20
Answer: B) 5 Solution: Use the Second Law of Thermodynamics to find the COP: COP = T2 / (T1 - T2). Common Wrong Answer: Option A, which is the ratio of heat removed to heat added.
Question 2: A heat engine operates between 100°C and 500°C. The efficiency of the engine is defined as the ratio of work done to heat added. What is the efficiency of the engine?
A) 20% B) 30% C) 40% D) 50%
Answer: C) 40% Solution: Use the First Law of Thermodynamics to find the efficiency: ? = 1 - (T1 / T2). Common Wrong Answer: Option B, which is the ratio of work done to heat removed.
Question 3: A system undergoes a process from state A to state B. The change in entropy is 10 J/K. If the temperature at state A is 300 K, what is the change in entropy at state B?
A) 5 J/K B) 10 J/K C) 15 J/K D) 20 J/K
Answer: B) 10 J/K Solution: Use the Second Law of Thermodynamics to find the change in entropy: ?S = ?Q / T. Common Wrong Answer: Option C, which is the sum of the change in entropy at state A and state B.
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