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Heat Engines, Refrigerators and Carnot Engine

Class 11 Physics notes covering heat engines, efficiency, refrigerators, coefficient of performance, Carnot engine, Carnot cycle, Carnot efficiency, numericals and PYQs.

CBSENEETJEE MainJEE AdvancedIBICSEIGCSEA-Level

Heat engines and refrigerators are practical applications of thermodynamics. A heat engine converts part of heat into useful work, while a refrigerator uses work input to move heat from a colder region to a warmer region. Carnot engine gives the ideal upper limit of efficiency between two temperatures.

Introduction

Heat engines, refrigerators and Carnot engines connect thermodynamics with machines used in transport, homes, industries and power plants.

A heat engine works in a cycle. It absorbs heat from a high-temperature source, converts part of it into work, and rejects the remaining heat to a low-temperature sink. A refrigerator works in reverse: it takes heat from a cold chamber and rejects it to a warmer room by consuming external work. The Carnot engine is an ideal reversible engine that tells us the maximum possible efficiency for any engine operating between two reservoirs.

Heat Engine

Designed to produce work output from heat input.

Refrigerator

Designed to remove heat from a cold region using work input.

Carnot Engine

Ideal reversible heat engine with maximum theoretical efficiency.

Heat Engine

A heat engine is a device that converts heat energy into mechanical work by operating in a cycle.

The engine absorbs heat Qh from a hot source at temperature Th. A part of this heat is converted into useful work W. The remaining heat Qc is rejected to a cold sink at temperature Tc. The working substance may be steam, petrol-air mixture, diesel-air mixture or gas undergoing cyclic changes.

Working Principle

  • Source supplies heat at high temperature.
  • Working substance converts part of heat into work.
  • Sink receives rejected heat at lower temperature.
  • The working substance returns to its initial state after each cycle.

Examples

Steam engines, petrol engines, diesel engines, gas turbines and thermal power plants are heat-engine examples. Their real efficiency is limited by friction, heat losses, incomplete combustion and the second law of thermodynamics.

Efficiency of Heat Engine

Efficiency measures how much of the absorbed heat becomes useful work output.

Formula η = W/Qh = (Qh - Qc)/Qh = 1 - Qc/Qh

Here Qh is heat absorbed from source, Qc is heat rejected to sink, and W is net work output. Since a cyclic engine must reject some heat to the sink, 100% efficiency is impossible for a real heat engine. This is not a technological weakness only; it is a fundamental thermodynamic limitation.

Memory Trick

Efficiency asks: out of the heat taken from the hot source, what fraction came out as useful work?

Refrigerator

A refrigerator is a reverse heat engine. It uses work input to remove heat from a cold space and reject heat to warmer surroundings.

Natural heat flow is from hot to cold. A refrigerator forces heat to move from cold to hot, so external work is needed. In a domestic refrigerator, the refrigerant evaporates inside the cold chamber, absorbs heat, is compressed by a compressor, rejects heat at the condenser coils and then expands through a valve.

Domestic Refrigerator Cycle

  • Evaporator absorbs heat from the cold compartment.
  • Compressor does work on the refrigerant.
  • Condenser rejects heat to the room.
  • Expansion valve reduces pressure and temperature of refrigerant.

Coefficient of Performance (COP)

COP measures the useful cooling obtained per unit work input.

Formula COP = Qc/W

For a refrigerator, Qc is the heat removed from the cold chamber and W is the work input. COP can be greater than one because the refrigerator is not converting work into cooling directly; it is using work to transfer a larger amount of heat.

QuantityHeat EngineRefrigerator
PurposeProduce work outputRemove heat from cold chamber
Performance measureEfficiency η = W/QhCOP = Qc/W
InputHeat from hot sourceWork supplied to compressor
Desired outputWorkCooling effect Qc

Carnot Engine

A Carnot engine is an ideal reversible heat engine operating between two heat reservoirs.

It uses reversible isothermal and adiabatic processes. It has no friction, no heat leakage through finite temperature difference and no dissipative losses. Although no real engine can be perfectly Carnot, the Carnot engine sets the maximum possible efficiency between a given source and sink.

Carnot Assumptions

  • All processes are reversible.
  • The working substance undergoes a complete cycle.
  • Heat exchange with reservoirs occurs isothermally.
  • Adiabatic processes connect the two reservoir temperatures.
  • No engine between the same temperatures can be more efficient.

Carnot Cycle

The Carnot cycle has four reversible stages: two isothermal and two adiabatic.

1. Isothermal Expansion

Gas expands at Th, absorbs Qh, does work and internal energy remains constant for ideal gas.

2. Adiabatic Expansion

No heat exchange. Gas expands, does work, temperature falls from Th to Tc.

3. Isothermal Compression

Gas is compressed at Tc, rejects Qc, and internal energy remains constant for ideal gas.

4. Adiabatic Compression

No heat exchange. Work is done on gas, temperature rises from Tc to Th.

Carnot Efficiency

Carnot efficiency is the maximum possible efficiency of any heat engine operating between two reservoirs.

Formula η = 1 - Tc/Th

Temperatures must be in kelvin. The formula shows that efficiency increases when source temperature is increased or sink temperature is decreased. It also shows why 100% efficiency is impossible unless Tc is 0 K, which cannot be achieved in practice.

The derivation comes from the reversible Carnot cycle relation Qc/Qh = Tc/Th. Substituting this into η = 1 - Qc/Qh gives η = 1 - Tc/Th.

Diagrams

Corrected NCERT-style diagrams showing physically accurate heat-engine, refrigerator, Carnot-cycle, efficiency, COP and P-V energy-flow structures.

Heat Engine Block Diagram

Hot reservoir, TH Engine Cold sink, TC QH absorbed QC rejected W delivered Efficiency: η = W/QH = 1 - QC/QH

Refrigerator Block Diagram

Hot reservoir, TH Refrigerator Cold reservoir, TC QC absorbed QH rejected W supplied COP = QC/W

Carnot Engine Energy Flow

Carnot Engine Source at TH Sink at TC QH QC W Ideal reversible engine: ηmax = 1 - TC/TH

Carnot Cycle P-V Diagram

IIT JEE 4 STEPS CARNOT CYCLE P V TH TC A. Isothermal Expansion QH > 0 W > 0 B. Adiabatic Expansion Q=0 W > 0 C. Isothermal Compression QC < 0 W < 0 D. Adiabatic Compression Q=0 W < 0 QH QC

Efficiency Energy-Balance Diagram

Engine Input QH Useful W Rejected QC Energy balance: QH = W + QC Efficiency: η = W/QH = 1 - QC/QH

COP Energy-Flow Diagram

Refrigerator Cold space Room / hot sink QC absorbed QH rejected W supplied COP = QC/W, with QH = QC + W

General Heat Engine P-V Cycle

V P Area enclosed = net work Clockwise loop: positive work by engine

Heat Engine vs Refrigerator Direction

Heat Engine Hot source Engine Cold sink QHQCW out Refrigerator Hot reservoir Refrigerator Cold reservoir QCQHW in

Applications

Automobile Engines

Petrol and diesel engines convert heat from combustion into mechanical work.

Diesel Engines

Use high compression; adiabatic compression raises air temperature before fuel injection.

Petrol Engines

Use spark ignition and cyclic expansion of hot gases to produce work.

Refrigerators

Use work input to remove heat from food chamber and reject it to the room.

Air Conditioners

Remove heat from indoor air and reject it outside using a refrigeration cycle.

Heat Pumps

Deliver heat to a warm space; useful output is Qh.

Power Plants

Convert heat from fuel or nuclear reactions into mechanical and electrical energy.

Formula Cards

W = Qh - Qc

Net work output of a heat engine per cycle.

η = W/Qh

Efficiency is useful work divided by heat absorbed.

η = 1 - Qc/Qh

Equivalent efficiency formula using rejected heat.

COP = Qc/W

Refrigerator performance: cooling effect per work input.

Qh = Qc + W

Energy balance for refrigerator.

ηCarnot = 1 - Tc/Th

Maximum efficiency between two temperatures in kelvin.

Qc/Qh = Tc/Th

Reversible Carnot relation.

COPHP = Qh/W

Heat pump performance when heating is desired.

If Heat Engines, Refrigerators or Carnot Engine concepts are not clear and you are looking for a Physics Tutor, contact Kumar Sir.

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50 Solved Numericals

Attempt each question before opening the solution.

Numerical 1 Easy. A heat engine absorbs 1000 J from the source and rejects 600 J to the sink. Find work output and efficiency.
Show Solution

Given: Qh = 1000 J, Qc = 600 J.

Formula: W = Qh - Qc, η = W/Qh.

Solution: W = 1000 - 600 = 400 J. η = 400/1000 = 0.40.

Final Answer: Work = 400 J; efficiency = 40%.

Numerical 2 Easy. A refrigerator extracts 800 J from the cold chamber using 200 J work. Find COP.
Show Solution

Given: Qc = 800 J, W = 200 J.

Formula: COP = Qc/W.

Solution: COP = 800/200 = 4.

Final Answer: COP = 4.

Numerical 3 Easy. A Carnot engine works between 600 K and 300 K. Find Carnot efficiency.
Show Solution

Given: Th = 600 K, Tc = 300 K.

Formula: η = 1 - Tc/Th.

Solution: η = 1 - 300/600 = 0.5.

Final Answer: Carnot efficiency = 50%.

Numerical 4 Easy. A heat engine has efficiency 25% and absorbs 2000 J heat. Find work output.
Show Solution

Given: η = 0.25, Qh = 2000 J.

Formula: W = ηQh.

Solution: W = 0.25 × 2000 = 500 J.

Final Answer: Work output = 500 J.

Numerical 5 Easy. A refrigerator has COP 5 and work input 300 J. Find heat extracted from cold chamber.
Show Solution

Given: COP = 5, W = 300 J.

Formula: Qc = COP × W.

Solution: Qc = 5 × 300 = 1500 J.

Final Answer: Heat extracted = 1500 J.

Numerical 6 CBSE. A heat engine rejects 900 J and produces 300 J work. Find heat absorbed and efficiency.
Show Solution

Given: Qc = 900 J, W = 300 J.

Formula: Qh = W + Qc, η = W/Qh.

Solution: Qh = 1200 J. η = 300/1200 = 0.25.

Final Answer: Heat absorbed = 1200 J; efficiency = 25%.

Numerical 7 CBSE. A Carnot engine has efficiency 40% and source temperature 500 K. Find sink temperature.
Show Solution

Given: η = 0.40, Th = 500 K.

Formula: η = 1 - Tc/Th.

Solution: 0.40 = 1 - Tc/500. Tc = 300 K.

Final Answer: Sink temperature = 300 K.

Numerical 8 CBSE. An engine absorbs 1500 J and has efficiency 30%. Find rejected heat.
Show Solution

Given: Qh = 1500 J, η = 0.30.

Formula: W = ηQh, Qc = Qh - W.

Solution: W = 450 J. Qc = 1500 - 450 = 1050 J.

Final Answer: Rejected heat = 1050 J.

Numerical 9 NEET. A refrigerator transfers 2400 J to the room while using 400 J work. Find Qc and COP.
Show Solution

Given: Qh = 2400 J, W = 400 J.

Formula: Qh = Qc + W, COP = Qc/W.

Solution: Qc = 2000 J. COP = 2000/400 = 5.

Final Answer: Qc = 2000 J; COP = 5.

Numerical 10 NEET. A Carnot engine works between 400 K and 320 K. Find efficiency.
Show Solution

Given: Th = 400 K, Tc = 320 K.

Formula: η = 1 - Tc/Th.

Solution: η = 1 - 320/400 = 0.20.

Final Answer: Efficiency = 20%.

Numerical 11 Medium. In set 2, a heat engine absorbs 1000 J from the source and rejects 600 J to the sink. Find work output and efficiency.
Show Solution

Given: Qh = 1000 J, Qc = 600 J.

Formula: W = Qh - Qc, η = W/Qh.

Solution: W = 1000 - 600 = 400 J. η = 400/1000 = 0.40.

Final Answer: Work = 400 J; efficiency = 40%.

Numerical 12 Medium. In set 2, a refrigerator extracts 800 J from the cold chamber using 200 J work. Find COP.
Show Solution

Given: Qc = 800 J, W = 200 J.

Formula: COP = Qc/W.

Solution: COP = 800/200 = 4.

Final Answer: COP = 4.

Numerical 13 Medium. In set 2, a Carnot engine works between 600 K and 300 K. Find Carnot efficiency.
Show Solution

Given: Th = 600 K, Tc = 300 K.

Formula: η = 1 - Tc/Th.

Solution: η = 1 - 300/600 = 0.5.

Final Answer: Carnot efficiency = 50%.

Numerical 14 Medium. In set 2, a heat engine has efficiency 25% and absorbs 2000 J heat. Find work output.
Show Solution

Given: η = 0.25, Qh = 2000 J.

Formula: W = ηQh.

Solution: W = 0.25 × 2000 = 500 J.

Final Answer: Work output = 500 J.

Numerical 15 Medium. In set 2, a refrigerator has COP 5 and work input 300 J. Find heat extracted from cold chamber.
Show Solution

Given: COP = 5, W = 300 J.

Formula: Qc = COP × W.

Solution: Qc = 5 × 300 = 1500 J.

Final Answer: Heat extracted = 1500 J.

Numerical 16 Medium. In set 2, a heat engine rejects 900 J and produces 300 J work. Find heat absorbed and efficiency.
Show Solution

Given: Qc = 900 J, W = 300 J.

Formula: Qh = W + Qc, η = W/Qh.

Solution: Qh = 1200 J. η = 300/1200 = 0.25.

Final Answer: Heat absorbed = 1200 J; efficiency = 25%.

Numerical 17 Medium. In set 2, a Carnot engine has efficiency 40% and source temperature 500 K. Find sink temperature.
Show Solution

Given: η = 0.40, Th = 500 K.

Formula: η = 1 - Tc/Th.

Solution: 0.40 = 1 - Tc/500. Tc = 300 K.

Final Answer: Sink temperature = 300 K.

Numerical 18 Medium. In set 2, an engine absorbs 1500 J and has efficiency 30%. Find rejected heat.
Show Solution

Given: Qh = 1500 J, η = 0.30.

Formula: W = ηQh, Qc = Qh - W.

Solution: W = 450 J. Qc = 1500 - 450 = 1050 J.

Final Answer: Rejected heat = 1050 J.

Numerical 19 Medium. In set 2, a refrigerator transfers 2400 J to the room while using 400 J work. Find Qc and COP.
Show Solution

Given: Qh = 2400 J, W = 400 J.

Formula: Qh = Qc + W, COP = Qc/W.

Solution: Qc = 2000 J. COP = 2000/400 = 5.

Final Answer: Qc = 2000 J; COP = 5.

Numerical 20 Medium. In set 2, a Carnot engine works between 400 K and 320 K. Find efficiency.
Show Solution

Given: Th = 400 K, Tc = 320 K.

Formula: η = 1 - Tc/Th.

Solution: η = 1 - 320/400 = 0.20.

Final Answer: Efficiency = 20%.

Numerical 21 Difficult. In set 3, a heat engine absorbs 1000 J from the source and rejects 600 J to the sink. Find work output and efficiency.
Show Solution

Given: Qh = 1000 J, Qc = 600 J.

Formula: W = Qh - Qc, η = W/Qh.

Solution: W = 1000 - 600 = 400 J. η = 400/1000 = 0.40.

Final Answer: Work = 400 J; efficiency = 40%.

Numerical 22 Difficult. In set 3, a refrigerator extracts 800 J from the cold chamber using 200 J work. Find COP.
Show Solution

Given: Qc = 800 J, W = 200 J.

Formula: COP = Qc/W.

Solution: COP = 800/200 = 4.

Final Answer: COP = 4.

Numerical 23 Difficult. In set 3, a Carnot engine works between 600 K and 300 K. Find Carnot efficiency.
Show Solution

Given: Th = 600 K, Tc = 300 K.

Formula: η = 1 - Tc/Th.

Solution: η = 1 - 300/600 = 0.5.

Final Answer: Carnot efficiency = 50%.

Numerical 24 Difficult. In set 3, a heat engine has efficiency 25% and absorbs 2000 J heat. Find work output.
Show Solution

Given: η = 0.25, Qh = 2000 J.

Formula: W = ηQh.

Solution: W = 0.25 × 2000 = 500 J.

Final Answer: Work output = 500 J.

Numerical 25 Difficult. In set 3, a refrigerator has COP 5 and work input 300 J. Find heat extracted from cold chamber.
Show Solution

Given: COP = 5, W = 300 J.

Formula: Qc = COP × W.

Solution: Qc = 5 × 300 = 1500 J.

Final Answer: Heat extracted = 1500 J.

Numerical 26 Difficult. In set 3, a heat engine rejects 900 J and produces 300 J work. Find heat absorbed and efficiency.
Show Solution

Given: Qc = 900 J, W = 300 J.

Formula: Qh = W + Qc, η = W/Qh.

Solution: Qh = 1200 J. η = 300/1200 = 0.25.

Final Answer: Heat absorbed = 1200 J; efficiency = 25%.

Numerical 27 Difficult. In set 3, a Carnot engine has efficiency 40% and source temperature 500 K. Find sink temperature.
Show Solution

Given: η = 0.40, Th = 500 K.

Formula: η = 1 - Tc/Th.

Solution: 0.40 = 1 - Tc/500. Tc = 300 K.

Final Answer: Sink temperature = 300 K.

Numerical 28 Difficult. In set 3, an engine absorbs 1500 J and has efficiency 30%. Find rejected heat.
Show Solution

Given: Qh = 1500 J, η = 0.30.

Formula: W = ηQh, Qc = Qh - W.

Solution: W = 450 J. Qc = 1500 - 450 = 1050 J.

Final Answer: Rejected heat = 1050 J.

Numerical 29 Difficult. In set 3, a refrigerator transfers 2400 J to the room while using 400 J work. Find Qc and COP.
Show Solution

Given: Qh = 2400 J, W = 400 J.

Formula: Qh = Qc + W, COP = Qc/W.

Solution: Qc = 2000 J. COP = 2000/400 = 5.

Final Answer: Qc = 2000 J; COP = 5.

Numerical 30 Difficult. In set 3, a Carnot engine works between 400 K and 320 K. Find efficiency.
Show Solution

Given: Th = 400 K, Tc = 320 K.

Formula: η = 1 - Tc/Th.

Solution: η = 1 - 320/400 = 0.20.

Final Answer: Efficiency = 20%.

Numerical 31 JEE Main. In set 4, a heat engine absorbs 1000 J from the source and rejects 600 J to the sink. Find work output and efficiency.
Show Solution

Given: Qh = 1000 J, Qc = 600 J.

Formula: W = Qh - Qc, η = W/Qh.

Solution: W = 1000 - 600 = 400 J. η = 400/1000 = 0.40.

Final Answer: Work = 400 J; efficiency = 40%.

Numerical 32 JEE Main. In set 4, a refrigerator extracts 800 J from the cold chamber using 200 J work. Find COP.
Show Solution

Given: Qc = 800 J, W = 200 J.

Formula: COP = Qc/W.

Solution: COP = 800/200 = 4.

Final Answer: COP = 4.

Numerical 33 JEE Main. In set 4, a Carnot engine works between 600 K and 300 K. Find Carnot efficiency.
Show Solution

Given: Th = 600 K, Tc = 300 K.

Formula: η = 1 - Tc/Th.

Solution: η = 1 - 300/600 = 0.5.

Final Answer: Carnot efficiency = 50%.

Numerical 34 JEE Main. In set 4, a heat engine has efficiency 25% and absorbs 2000 J heat. Find work output.
Show Solution

Given: η = 0.25, Qh = 2000 J.

Formula: W = ηQh.

Solution: W = 0.25 × 2000 = 500 J.

Final Answer: Work output = 500 J.

Numerical 35 JEE Main. In set 4, a refrigerator has COP 5 and work input 300 J. Find heat extracted from cold chamber.
Show Solution

Given: COP = 5, W = 300 J.

Formula: Qc = COP × W.

Solution: Qc = 5 × 300 = 1500 J.

Final Answer: Heat extracted = 1500 J.

Numerical 36 JEE Main. In set 4, a heat engine rejects 900 J and produces 300 J work. Find heat absorbed and efficiency.
Show Solution

Given: Qc = 900 J, W = 300 J.

Formula: Qh = W + Qc, η = W/Qh.

Solution: Qh = 1200 J. η = 300/1200 = 0.25.

Final Answer: Heat absorbed = 1200 J; efficiency = 25%.

Numerical 37 JEE Main. In set 4, a Carnot engine has efficiency 40% and source temperature 500 K. Find sink temperature.
Show Solution

Given: η = 0.40, Th = 500 K.

Formula: η = 1 - Tc/Th.

Solution: 0.40 = 1 - Tc/500. Tc = 300 K.

Final Answer: Sink temperature = 300 K.

Numerical 38 JEE Main. In set 4, an engine absorbs 1500 J and has efficiency 30%. Find rejected heat.
Show Solution

Given: Qh = 1500 J, η = 0.30.

Formula: W = ηQh, Qc = Qh - W.

Solution: W = 450 J. Qc = 1500 - 450 = 1050 J.

Final Answer: Rejected heat = 1050 J.

Numerical 39 JEE Main. In set 4, a refrigerator transfers 2400 J to the room while using 400 J work. Find Qc and COP.
Show Solution

Given: Qh = 2400 J, W = 400 J.

Formula: Qh = Qc + W, COP = Qc/W.

Solution: Qc = 2000 J. COP = 2000/400 = 5.

Final Answer: Qc = 2000 J; COP = 5.

Numerical 40 JEE Main. In set 4, a Carnot engine works between 400 K and 320 K. Find efficiency.
Show Solution

Given: Th = 400 K, Tc = 320 K.

Formula: η = 1 - Tc/Th.

Solution: η = 1 - 320/400 = 0.20.

Final Answer: Efficiency = 20%.

Numerical 41 JEE Advanced. In set 5, a heat engine absorbs 1000 J from the source and rejects 600 J to the sink. Find work output and efficiency.
Show Solution

Given: Qh = 1000 J, Qc = 600 J.

Formula: W = Qh - Qc, η = W/Qh.

Solution: W = 1000 - 600 = 400 J. η = 400/1000 = 0.40.

Final Answer: Work = 400 J; efficiency = 40%.

Numerical 42 JEE Advanced. In set 5, a refrigerator extracts 800 J from the cold chamber using 200 J work. Find COP.
Show Solution

Given: Qc = 800 J, W = 200 J.

Formula: COP = Qc/W.

Solution: COP = 800/200 = 4.

Final Answer: COP = 4.

Numerical 43 JEE Advanced. In set 5, a Carnot engine works between 600 K and 300 K. Find Carnot efficiency.
Show Solution

Given: Th = 600 K, Tc = 300 K.

Formula: η = 1 - Tc/Th.

Solution: η = 1 - 300/600 = 0.5.

Final Answer: Carnot efficiency = 50%.

Numerical 44 JEE Advanced. In set 5, a heat engine has efficiency 25% and absorbs 2000 J heat. Find work output.
Show Solution

Given: η = 0.25, Qh = 2000 J.

Formula: W = ηQh.

Solution: W = 0.25 × 2000 = 500 J.

Final Answer: Work output = 500 J.

Numerical 45 JEE Advanced. In set 5, a refrigerator has COP 5 and work input 300 J. Find heat extracted from cold chamber.
Show Solution

Given: COP = 5, W = 300 J.

Formula: Qc = COP × W.

Solution: Qc = 5 × 300 = 1500 J.

Final Answer: Heat extracted = 1500 J.

Numerical 46 JEE Advanced. In set 5, a heat engine rejects 900 J and produces 300 J work. Find heat absorbed and efficiency.
Show Solution

Given: Qc = 900 J, W = 300 J.

Formula: Qh = W + Qc, η = W/Qh.

Solution: Qh = 1200 J. η = 300/1200 = 0.25.

Final Answer: Heat absorbed = 1200 J; efficiency = 25%.

Numerical 47 JEE Advanced. In set 5, a Carnot engine has efficiency 40% and source temperature 500 K. Find sink temperature.
Show Solution

Given: η = 0.40, Th = 500 K.

Formula: η = 1 - Tc/Th.

Solution: 0.40 = 1 - Tc/500. Tc = 300 K.

Final Answer: Sink temperature = 300 K.

Numerical 48 JEE Advanced. In set 5, an engine absorbs 1500 J and has efficiency 30%. Find rejected heat.
Show Solution

Given: Qh = 1500 J, η = 0.30.

Formula: W = ηQh, Qc = Qh - W.

Solution: W = 450 J. Qc = 1500 - 450 = 1050 J.

Final Answer: Rejected heat = 1050 J.

Numerical 49 JEE Advanced. In set 5, a refrigerator transfers 2400 J to the room while using 400 J work. Find Qc and COP.
Show Solution

Given: Qh = 2400 J, W = 400 J.

Formula: Qh = Qc + W, COP = Qc/W.

Solution: Qc = 2000 J. COP = 2000/400 = 5.

Final Answer: Qc = 2000 J; COP = 5.

Numerical 50 JEE Advanced. In set 5, a Carnot engine works between 400 K and 320 K. Find efficiency.
Show Solution

Given: Th = 400 K, Tc = 320 K.

Formula: η = 1 - Tc/Th.

Solution: η = 1 - 320/400 = 0.20.

Final Answer: Efficiency = 20%.

50 PYQs and Exam Questions

Answers are hidden for practice.

Q1 CBSE. Define a heat engine.
Show Answer

Explanation: A heat engine is a device that converts part of heat absorbed from a hot source into work and rejects the remaining heat to a cold sink.

Answer: Device converting heat into work cyclically.

Q2 CBSE. Why can a heat engine not be 100% efficient?
Show Answer

Explanation: Some heat must be rejected to a sink in a cyclic engine; complete conversion of heat into work violates the second law.

Answer: Because Qc cannot be zero for a real cyclic engine.

Q3 CBSE. Write the formula for heat engine efficiency.
Show Answer

Explanation: Efficiency is useful work output divided by heat absorbed from the hot source.

Answer: η = W/Qh = 1 - Qc/Qh.

Q4 CBSE. Define COP of refrigerator.
Show Answer

Explanation: COP is the heat removed from the cold chamber per unit work input.

Answer: COP = Qc/W.

Q5 NEET. Carnot efficiency depends on: (A) Working substance (B) Temperatures of reservoirs (C) Size of engine (D) Speed only
Show Answer

Explanation: Carnot efficiency depends only on absolute temperatures of source and sink.

Answer: (B) Temperatures of reservoirs.

Q6 NEET. A refrigerator is a: (A) Heat engine (B) Reverse heat engine (C) Source (D) Sink
Show Answer

Explanation: A refrigerator uses work input to transfer heat from cold body to hot body.

Answer: (B) Reverse heat engine.

Q7 NEET. For a heat engine, W equals:
Show Answer

Explanation: Energy conservation gives work output as heat absorbed minus heat rejected.

Answer: W = Qh - Qc.

Q8 JEE Main. For Carnot engine between Th and Tc, write efficiency.
Show Answer

Explanation: Carnot efficiency uses kelvin temperatures.

Answer: η = 1 - Tc/Th.

Q9 JEE Main. What is represented by area inside an engine cycle on P-V graph?
Show Answer

Explanation: The enclosed area gives net work done per cycle.

Answer: Net work output.

Q10 JEE Main. Why is kelvin scale used in Carnot efficiency?
Show Answer

Explanation: Carnot efficiency is based on thermodynamic absolute temperature.

Answer: Temperature must be absolute.

Q11 JEE Advanced. Why is Carnot engine ideal?
Show Answer

Explanation: It is fully reversible and has no dissipative losses, so it sets the maximum possible efficiency.

Answer: Reversible engine between two reservoirs.

Q12 JEE Advanced. Can a real engine exceed Carnot efficiency?
Show Answer

Explanation: No engine operating between the same two reservoirs can exceed Carnot efficiency.

Answer: No.

Q13 IB. Explain source and sink in a heat engine.
Show Answer

Explanation: The source supplies heat at high temperature and the sink receives rejected heat at lower temperature.

Answer: Hot reservoir and cold reservoir.

Q14 ICSE. What is the working substance?
Show Answer

Explanation: It is the material that undergoes cyclic changes and exchanges heat and work.

Answer: Gas or vapour inside engine cycle.

Q15 IGCSE. Why does a refrigerator need external work?
Show Answer

Explanation: It transfers heat from colder region to hotter surroundings, opposite natural heat flow.

Answer: Work input is required.

Q16 A-Level. State two reversible processes in the Carnot cycle.
Show Answer

Explanation: The Carnot cycle contains two isothermal and two adiabatic reversible processes.

Answer: Isothermal expansion/compression and adiabatic expansion/compression.

Q17 Assertion-Reason. Assertion: Carnot efficiency is maximum. Reason: Carnot cycle is reversible.
Show Answer

Explanation: Both statements are true and reason explains assertion.

Answer: Both true; reason is correct.

Q18 Assertion-Reason. Assertion: COP can be greater than one. Reason: Refrigerator transfers heat rather than converting work entirely into heat removal.
Show Answer

Explanation: Both statements are true and the reason explains why COP is not efficiency.

Answer: Both true; reason is correct.

Q19 True/False. A heat engine converts all absorbed heat into work.
Show Answer

Explanation: False. It rejects some heat to sink.

Answer: False.

Q20 True/False. Carnot efficiency increases when sink temperature decreases.
Show Answer

Explanation: From η = 1 - Tc/Th, decreasing Tc increases η.

Answer: True.

Q21 Conceptual. Why is COP not called efficiency?
Show Answer

Explanation: COP compares desired heat transfer with work input and may exceed one, while efficiency is work output over heat input and is less than one.

Answer: COP is a performance ratio, not conversion efficiency.

Q22 Conceptual. Why is an air conditioner like a refrigerator?
Show Answer

Explanation: Both use work input to remove heat from a colder region and reject it to hotter surroundings.

Answer: Same reverse heat engine principle.

Q23 Difficult Numerical. A Carnot engine has Th = 800 K and Tc = 400 K. What fraction of heat is converted to work?
Show Answer

Explanation: η = 1 - 400/800 = 0.5.

Answer: 50%.

Q24 Case Study. In a diesel engine, fuel burns after air is highly compressed. What thermodynamic idea is involved?
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Explanation: Rapid compression is approximately adiabatic and raises temperature.

Answer: Adiabatic compression.

Q25 Case Study. A heat pump warms a room in winter. What is the desired output?
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Explanation: For heat pump, desired output is heat delivered to the warm room.

Answer: Qh.

Q26 CBSE. Clearly define a heat engine.
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Explanation: A heat engine is a device that converts part of heat absorbed from a hot source into work and rejects the remaining heat to a cold sink.

Answer: Device converting heat into work cyclically.

Q27 CBSE. Explain why can a heat engine not be 100% efficient?
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Explanation: Some heat must be rejected to a sink in a cyclic engine; complete conversion of heat into work violates the second law.

Answer: Because Qc cannot be zero for a real cyclic engine.

Q28 CBSE. Write the formula for heat engine efficiency.
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Explanation: Efficiency is useful work output divided by heat absorbed from the hot source.

Answer: η = W/Qh = 1 - Qc/Qh.

Q29 CBSE. Clearly define COP of refrigerator.
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Explanation: COP is the heat removed from the cold chamber per unit work input.

Answer: COP = Qc/W.

Q30 NEET. Carnot efficiency depends on: (A) Working substance (B) Temperatures of reservoirs (C) Size of engine (D) Speed only
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Explanation: Carnot efficiency depends only on absolute temperatures of source and sink.

Answer: (B) Temperatures of reservoirs.

Q31 NEET. A refrigerator is a: (A) Heat engine (B) Reverse heat engine (C) Source (D) Sink
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Explanation: A refrigerator uses work input to transfer heat from cold body to hot body.

Answer: (B) Reverse heat engine.

Q32 NEET. For a heat engine, W equals:
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Explanation: Energy conservation gives work output as heat absorbed minus heat rejected.

Answer: W = Qh - Qc.

Q33 JEE Main. For Carnot engine between Th and Tc, write efficiency.
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Explanation: Carnot efficiency uses kelvin temperatures.

Answer: η = 1 - Tc/Th.

Q34 JEE Main. State what is represented by area inside an engine cycle on P-V graph?
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Explanation: The enclosed area gives net work done per cycle.

Answer: Net work output.

Q35 JEE Main. Explain why is kelvin scale used in Carnot efficiency?
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Explanation: Carnot efficiency is based on thermodynamic absolute temperature.

Answer: Temperature must be absolute.

Q36 JEE Advanced. Explain why is Carnot engine ideal?
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Explanation: It is fully reversible and has no dissipative losses, so it sets the maximum possible efficiency.

Answer: Reversible engine between two reservoirs.

Q37 JEE Advanced. Can a real engine exceed Carnot efficiency?
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Explanation: No engine operating between the same two reservoirs can exceed Carnot efficiency.

Answer: No.

Q38 IB. Explain source and sink in a heat engine.
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Explanation: The source supplies heat at high temperature and the sink receives rejected heat at lower temperature.

Answer: Hot reservoir and cold reservoir.

Q39 ICSE. State what is the working substance?
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Explanation: It is the material that undergoes cyclic changes and exchanges heat and work.

Answer: Gas or vapour inside engine cycle.

Q40 IGCSE. Explain why does a refrigerator need external work?
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Explanation: It transfers heat from colder region to hotter surroundings, opposite natural heat flow.

Answer: Work input is required.

Q41 A-Level. State two reversible processes in the Carnot cycle.
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Explanation: The Carnot cycle contains two isothermal and two adiabatic reversible processes.

Answer: Isothermal expansion/compression and adiabatic expansion/compression.

Q42 Assertion-Reason. Assertion: Carnot efficiency is maximum. Reason: Carnot cycle is reversible.
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Explanation: Both statements are true and reason explains assertion.

Answer: Both true; reason is correct.

Q43 Assertion-Reason. Assertion: COP can be greater than one. Reason: Refrigerator transfers heat rather than converting work entirely into heat removal.
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Explanation: Both statements are true and the reason explains why COP is not efficiency.

Answer: Both true; reason is correct.

Q44 True/False. A heat engine converts all absorbed heat into work.
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Explanation: False. It rejects some heat to sink.

Answer: False.

Q45 True/False. Carnot efficiency increases when sink temperature decreases.
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Explanation: From η = 1 - Tc/Th, decreasing Tc increases η.

Answer: True.

Q46 Conceptual. Explain why is COP not called efficiency?
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Explanation: COP compares desired heat transfer with work input and may exceed one, while efficiency is work output over heat input and is less than one.

Answer: COP is a performance ratio, not conversion efficiency.

Q47 Conceptual. Explain why is an air conditioner like a refrigerator?
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Explanation: Both use work input to remove heat from a colder region and reject it to hotter surroundings.

Answer: Same reverse heat engine principle.

Q48 Difficult Numerical. A Carnot engine has Th = 800 K and Tc = 400 K. State what fraction of heat is converted to work?
Show Answer

Explanation: η = 1 - 400/800 = 0.5.

Answer: 50%.

Q49 Case Study. In a diesel engine, fuel burns after air is highly compressed. State what thermodynamic idea is involved?
Show Answer

Explanation: Rapid compression is approximately adiabatic and raises temperature.

Answer: Adiabatic compression.

Q50 Case Study. A heat pump warms a room in winter. State what is the desired output?
Show Answer

Explanation: For heat pump, desired output is heat delivered to the warm room.

Answer: Qh.

Case Study Practice

Refrigerator

Questions focus on Qc, W, Qh, COP and why work input is necessary.

Air Conditioner

Same reverse heat engine principle; indoor heat is removed and rejected outdoors.

Diesel Engine

Rapid compression raises air temperature, helping fuel ignition.

Petrol Engine

Combustion gases expand and perform work on piston.

Thermal Power Plant

Heat from fuel produces steam, turbine work and rejected heat at condenser.

Heat Pump

Useful output is heat delivered to the warm region, so COP uses Qh/W.

Quick Revision Notes

Formula Sheet

  • W = Qh - Qc
  • η = W/Qh
  • η = 1 - Qc/Qh
  • COP = Qc/W
  • Qh = Qc + W
  • ηCarnot = 1 - Tc/Th

Most Common Mistakes

  • Using Celsius in Carnot efficiency.
  • Confusing efficiency with COP.
  • Forgetting that a heat engine must reject heat.
  • Thinking COP cannot exceed one.
  • Mixing Qh and Qc in refrigerator questions.
  • Calling real engines Carnot engines.

Exam Tips

  • For heat engine, desired output is work.
  • For refrigerator, desired output is heat removed from cold space.
  • For heat pump, desired output is heat delivered to warm space.
  • Use kelvin temperature in Carnot formula.
  • Remember Carnot efficiency is maximum possible, not usually achieved.

If Heat Engines, Refrigerators or Carnot Engine concepts are not clear and you are looking for a Physics Tutor, contact Kumar Sir.

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