Vidaara.orgClass 11 · Physics
CodeVID-P11-11-SLE-01
Second Law, Heat Engines & Refrigerators — Assignment
Name: ____________________
Roll No.: __________
Date: ____________
General Instructions
- All questions are compulsory.
- Section A carries 1 mark each, Section B 2 marks, Section C 3 marks and Section D 5 marks.
- Show all working for Sections B, C and D. Only final answers are given at the end — for full solutions, raise your doubts with your teacher.
Section A — Multiple Choice Questions
5 × 1 = 5 marks
1.
The Clausius statement concerns the natural flow of:
- A.work
- B.heat from cold to hot
- C.charge
- D.momentum
2.
The work done by a heat engine per cycle is:
- A.$Q_1+Q_2$
- B.$Q_1-Q_2$
- C.$Q_2-Q_1$
- D.$Q_1 Q_2$
3.
A Carnot engine's efficiency depends only on:
- A.the working substance
- B.the source and sink temperatures
- C.the volume of gas
- D.the pressure
4.
The efficiency of a real heat engine is always:
- A.100%
- B.greater than the Carnot value
- C.less than the Carnot value
- D.exactly 50%
5.
The coefficient of performance of a refrigerator can be:
- A.only less than 1
- B.only equal to 1
- C.greater than 1
- D.negative
Section B — Short Answer (2 marks)
3 × 2 = 6 marks
6.
State the Kelvin–Planck statement of the second law.
7.
A Carnot engine works between 600 K and 300 K. Find its efficiency.
8.
Why can the coefficient of performance of a refrigerator exceed 1?
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
9.
A heat engine absorbs 800 J and rejects 600 J per cycle. Find the work done and efficiency.
10.
Describe the four strokes of the Carnot cycle.
Section D — Long Answer (5 marks)
1 × 5 = 5 marks
11.
Describe the working of a heat engine and derive the efficiency of a Carnot engine $\eta=1-\frac{T_2}{T_1}$. State why no engine can be 100% efficient.
Answer Key
Section A — Multiple Choice Questions
- (B) heat from cold to hot
- (B) $Q_1-Q_2$
- (B) the source and sink temperatures
- (C) less than the Carnot value
- (C) greater than 1
Section B — Short Answer (2 marks)
- No engine working in a cycle can take heat from a single reservoir and convert it entirely into work; some heat must always be rejected.
- $\eta=1-\frac{300}{600}=0.5=50\%$.
- Because it measures heat removed per unit work, $\beta=\frac{Q_2}{W}$, and $Q_2$ can be larger than $W$.
Section C — Short Answer (3 marks)
- $W=800-600=200\ \text{J}$; $\eta=1-\frac{600}{800}=0.25=25\%$.
- Isothermal expansion (heat $Q_1$ in at $T_1$), adiabatic expansion (cools to $T_2$), isothermal compression (heat $Q_2$ out at $T_2$), adiabatic compression (back to start).
Section D — Long Answer (5 marks)
- An engine absorbs $Q_1$ at $T_1$, does work $W=Q_1-Q_2$ and rejects $Q_2$ at $T_2$, so $\eta=1-\frac{Q_2}{Q_1}$. For the reversible Carnot cycle $\frac{Q_2}{Q_1}=\frac{T_2}{T_1}$, giving $\eta=1-\frac{T_2}{T_1}$. Since $T_2=0\ \text{K}$ is unattainable (and $Q_2>0$ by Kelvin–Planck), $\eta<1$ always.
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