Heat Engines and Carnot Cycle

Real-Life Hook: Your Car is a Heat Engine on Wheels!

Every time you start your car, you’re operating a sophisticated heat engine! The gasoline explodes at ~2000°C (high temperature), pushes pistons to do work (driving wheels), and exhausts waste heat through the radiator (low temperature).

Your car’s efficiency? Only about 25-30%! Where does the rest go?

• ~35% lost through exhaust
• ~30% lost through cooling system
• ~5% lost to friction

Even the best Formula 1 engines achieve only ~50% efficiency. Why can’t we do better? The Second Law of Thermodynamics sets a fundamental limit: the Carnot efficiency, which depends only on operating temperatures.

Understanding heat engines explains:

• Why hybrid cars are more efficient
• Why diesel engines beat petrol engines
• How power plants generate electricity
• Why we can’t build perpetual motion machines

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What is a Heat Engine?

Definition

Heat engine: A device that converts heat into mechanical work by operating in a cycle.

Basic Components

1. Hot reservoir (Source) at temperature $T_H$
2. Working substance (gas, steam, etc.)
3. Cold reservoir (Sink) at temperature $T_C$
4. Mechanical output (piston, turbine, etc.)

Operating Principle

In one complete cycle:

1. Absorb heat $Q_H$ from hot reservoir
2. Convert part to work $W$
3. Reject remaining heat $Q_C$ to cold reservoir
4. Return to initial state

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Energy Flow in Heat Engines

First Law Application

For complete cycle: $\Delta U = 0$ (returns to same state)

Efficiency

Why Efficiency Matters

Example: 1000 MW power plant with η = 40%

• Fuel energy input: 2500 MW
• Electrical output: 1000 MW (40%)
• Waste heat: 1500 MW (60%)
• This waste heat warms rivers, requires cooling towers!

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The Carnot Cycle: The Ideal Heat Engine

What is the Carnot Cycle?

The Carnot cycle is a theoretical ideal cycle consisting of:

1. Two isothermal processes (at $T_H$ and $T_C$)
2. Two adiabatic processes (connecting the isothermals)

All processes are reversible (quasi-static, frictionless).

The Four Steps

Step 1: Isothermal Expansion (A → B) at $T_H$

• Gas in contact with hot reservoir at $T_H$
• Expands from $V_A$ to $V_B$ at constant temperature
• Heat absorbed: $Q_H = nRT_H\ln(V_B/V_A)$
• Work done: $W_1 = Q_H$ (since $\Delta U = 0$)

Step 2: Adiabatic Expansion (B → C)

• Thermally isolated (no heat exchange)
• Expands from $V_B$ to $V_C$
• Temperature drops from $T_H$ to $T_C$
• Work done: $W_2 = nC_V(T_H - T_C)$
• Heat: $Q = 0$

Step 3: Isothermal Compression (C → D) at $T_C$

• Gas in contact with cold reservoir at $T_C$
• Compresses from $V_C$ to $V_D$ at constant temperature
• Heat rejected: $Q_C = nRT_C\ln(V_C/V_D)$
• Work done: $W_3 = -Q_C$ (negative, compression)

Step 4: Adiabatic Compression (D → A)

• Thermally isolated
• Compresses from $V_D$ to $V_A$
• Temperature rises from $T_C$ to $T_H$
• Work done: $W_4 = -nC_V(T_H - T_C)$
• Heat: $Q = 0$

Carnot Cycle on PV Diagram

Shape: Elongated loop with:

• Two hyperbolic curves (isothermals)
• Two steeper curves (adiabatics)
• Clockwise direction (heat engine)

Area enclosed = Net work done

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Carnot Efficiency: The Ultimate Limit

Derivation

Heat absorbed at $T_H$:

$$Q_H = nRT_H\ln\frac{V_B}{V_A}$$

Heat rejected at $T_C$:

$$Q_C = nRT_C\ln\frac{V_C}{V_D}$$

For adiabatic processes:

• B → C: $T_HV_B^{\gamma-1} = T_CV_C^{\gamma-1}$
• D → A: $T_CV_D^{\gamma-1} = T_HV_A^{\gamma-1}$

Dividing these:

$$\frac{V_B^{\gamma-1}}{V_D^{\gamma-1}} = \frac{V_C^{\gamma-1}}{V_A^{\gamma-1}}$$ $$\left(\frac{V_B}{V_D}\right)^{\gamma-1} = \left(\frac{V_C}{V_A}\right)^{\gamma-1}$$ $$\frac{V_B}{V_D} = \frac{V_C}{V_A}$$

Or: $\frac{V_B}{V_A} = \frac{V_C}{V_D}$

Therefore:

$$\frac{Q_C}{Q_H} = \frac{nRT_C\ln(V_C/V_D)}{nRT_H\ln(V_B/V_A)} = \frac{T_C}{T_H}$$

Carnot efficiency:

Implications

1. To increase efficiency:

   • Increase $T_H$ (hotter combustion)
   • Decrease $T_C$ (better cooling)

2. 100% efficiency requires:

   • $T_C = 0$ (absolute zero - impossible!)
   • Or $T_H = \infty$ (infinite temperature - impossible!)

3. Practical limits:

   • Car engines: $T_H ≈ 2000$ K, $T_C ≈ 350$ K → $\eta_{max} ≈ 82.5%$
   • Actual: ~30% (due to irreversibilities)

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Carnot’s Theorem

Statement

No heat engine operating between two reservoirs can be more efficient than a Carnot engine operating between the same two temperatures.

Corollaries

1. All reversible engines between same $T_H$ and $T_C$ have same efficiency (Carnot)
2. All irreversible engines have lower efficiency
3. Efficiency depends only on temperatures, not working substance

Why Carnot is Ideal

Reversible means:

• Quasi-static (infinitely slow)
• Frictionless
• No heat flow across finite ΔT
• Impossible to achieve in practice!

All real engines are irreversible → Less efficient than Carnot

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Real Heat Engines

1. Otto Cycle (Petrol Engine)

Processes:

1. Adiabatic compression (compression stroke)
2. Isochoric heat addition (spark ignition)
3. Adiabatic expansion (power stroke)
4. Isochoric heat rejection (exhaust)

Efficiency:

$$\eta_{Otto} = 1 - \frac{1}{r^{\gamma-1}}$$

where $r = V_{max}/V_{min}$ is compression ratio

Typical values:

• Compression ratio: 8-11
• Efficiency: 25-30% (real), 50-60% (theoretical)

Why less than Carnot?

• Isochoric processes (not isothermal)
• Friction, heat loss
• Incomplete combustion

2. Diesel Cycle

Processes:

1. Adiabatic compression (higher than Otto)
2. Isobaric heat addition (fuel injection)
3. Adiabatic expansion
4. Isochoric heat rejection

Efficiency:

$$\eta_{Diesel} = 1 - \frac{1}{r^{\gamma-1}} \times \frac{\rho^\gamma - 1}{\gamma(\rho - 1)}$$

where $\rho$ is cutoff ratio = $V_3/V_2$

Typical values:

• Compression ratio: 14-25 (higher than Otto!)
• Efficiency: 30-40% (real), 60-70% (theoretical)

Why more efficient than Otto?

• Higher compression ratio
• Higher operating temperatures

3. Stirling Cycle

Processes:

1. Isothermal expansion at $T_H$
2. Isochoric cooling
3. Isothermal compression at $T_C$
4. Isochoric heating

Efficiency:

$$\eta_{Stirling} = \eta_{Carnot} = 1 - \frac{T_C}{T_H}$$

Why not used more?

• Difficult to implement
• Low power output
• Slow speed
• Used in specialized applications (submarines, space)

4. Brayton Cycle (Gas Turbine)

Processes:

1. Adiabatic compression (compressor)
2. Isobaric heat addition (combustor)
3. Adiabatic expansion (turbine)
4. Isobaric heat rejection (exhaust)

Used in:

• Jet engines
• Power plants (combined cycle)

Efficiency:

$$\eta_{Brayton} = 1 - \frac{1}{r_p^{(\gamma-1)/\gamma}}$$

where $r_p = P_2/P_1$ is pressure ratio

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Increasing Efficiency: Practical Methods

1. Combined Cycle Power Plants

Concept: Combine gas turbine (Brayton) with steam turbine (Rankine)

How it works:

1. Gas turbine burns fuel → electricity (efficiency ~40%)
2. Exhaust heat boils water → steam turbine → more electricity
3. Combined efficiency: 55-60% (best for large plants!)

2. Cogeneration (CHP)

Combined Heat and Power: Use waste heat for heating/cooling

Overall utilization: Up to 90% (not efficiency, but energy use)

3. Higher Operating Temperatures

Materials challenge:

• Higher $T_H$ → Better efficiency
• But materials must withstand heat
• Modern turbines: Use ceramic coatings, active cooling

4. Regeneration and Recuperation

Concept: Use exhaust heat to preheat incoming air/fuel

Benefit: Less fuel needed for same power

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Common Mistakes & Tricks

Mistake 1: Using °C in Carnot Formula

❌ $\eta = 1 - 300/500$ (if temperatures in °C) ✅ Convert to Kelvin first!

Always use absolute temperature (Kelvin) in thermodynamic formulas.

Mistake 2: Thinking Real Engine Can Match Carnot

❌ Car engine between 2000 K and 300 K → η should be 85% ✅ Real engines: friction, heat loss, finite-speed → Much lower!

Carnot is theoretical limit, never achieved in practice.

Mistake 3: Confusing Heat and Temperature

• High temperature ≠ High heat content
• Carnot efficiency depends on temperature ratio, not heat amounts

Mistake 4: Reversible = Realistic

❌ Carnot is reversible, so we can build it ✅ Reversible requires infinitely slow, frictionless - impossible!

All real processes are irreversible.

Memory Trick: “Hot-Cold Gap”

$$\eta = 1 - \frac{T_C}{T_H} = \frac{T_H - T_C}{T_H}$$

Bigger the gap ($T_H - T_C$), better the efficiency!

Trick: Compression Ratio

Otto cycle: Higher compression → Better efficiency But: Too high → Knocking (pre-ignition)

Diesel: Can use higher compression (no premix, no knocking) → More efficient than petrol!

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Practice Problems

Level 1: JEE Main Basics

Q1. A Carnot engine operates between 400 K and 300 K. Find its efficiency and work done if 1000 J is absorbed from hot reservoir.

Solution

Efficiency:

$$\eta = 1 - \frac{T_C}{T_H} = 1 - \frac{300}{400} = 1 - 0.75 = 0.25 = 25\%$$

Work done:

$$W = \eta \times Q_H = 0.25 \times 1000 = 250 \text{ J}$$

Heat rejected:

$$Q_C = Q_H - W = 1000 - 250 = 750 \text{ J}$$

Q2. An engine has efficiency 40% and does 200 J of work per cycle. Find: (a) Heat absorbed (b) Heat rejected

Solution

(a) Heat absorbed:

$$\eta = \frac{W}{Q_H}$$ $$Q_H = \frac{W}{\eta} = \frac{200}{0.4} = 500 \text{ J}$$

(b) Heat rejected:

$$Q_C = Q_H - W = 500 - 200 = 300 \text{ J}$$

Or: $Q_C = Q_H(1 - \eta) = 500 \times 0.6 = 300$ J ✓

Q3. At what temperature is Carnot efficiency 50% if cold reservoir is at 27°C?

Solution

Given: $\eta = 0.5$, $T_C = 27°C = 300$ K

$$\eta = 1 - \frac{T_C}{T_H}$$ $$0.5 = 1 - \frac{300}{T_H}$$ $$\frac{300}{T_H} = 0.5$$ $$T_H = \frac{300}{0.5} = 600 \text{ K} = 327°C$$

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Level 2: JEE Main/Advanced

Q4. A Carnot engine working between 400 K and 300 K absorbs 600 J from hot reservoir. A second Carnot engine works between 300 K and 200 K, taking heat rejected by first engine. Find: (a) Work done by each engine (b) Total work (c) Overall efficiency if treated as single engine

Solution

Engine 1: Between 400 K and 300 K

$$\eta_1 = 1 - \frac{300}{400} = 0.25$$ $$W_1 = \eta_1 \times Q_{H1} = 0.25 \times 600 = 150 \text{ J}$$ $$Q_{C1} = Q_{H1} - W_1 = 600 - 150 = 450 \text{ J}$$

Engine 2: Between 300 K and 200 K

Heat input to engine 2: $Q_{H2} = Q_{C1} = 450$ J

$$\eta_2 = 1 - \frac{200}{300} = \frac{1}{3} ≈ 0.333$$ $$W_2 = \eta_2 \times Q_{H2} = \frac{1}{3} \times 450 = 150 \text{ J}$$

(b) Total work:

$$W_{total} = W_1 + W_2 = 150 + 150 = 300 \text{ J}$$

(c) Overall efficiency: Heat absorbed from 400 K reservoir: 600 J Total work: 300 J

$$\eta_{overall} = \frac{300}{600} = 0.5 = 50\%$$

Check using Carnot formula:

$$\eta_{overall} = 1 - \frac{200}{400} = 0.5$$

✓

Note: Two Carnot engines in series = One Carnot engine from highest to lowest temperature!

Q5. An Otto cycle has compression ratio 8 and operates with air (γ = 1.4). Find: (a) Theoretical efficiency (b) Compare with Carnot efficiency if $T_{max} = 2000$ K, $T_{min} = 300$ K

Solution

(a) Otto efficiency:

$$\eta_{Otto} = 1 - \frac{1}{r^{\gamma-1}}$$ $$\eta_{Otto} = 1 - \frac{1}{8^{1.4-1}} = 1 - \frac{1}{8^{0.4}}$$

Calculate: $8^{0.4} = (2^3)^{0.4} = 2^{1.2} = 2.297$

$$\eta_{Otto} = 1 - \frac{1}{2.297} = 1 - 0.435 = 0.565 = 56.5\%$$

(b) Carnot efficiency:

$$\eta_{Carnot} = 1 - \frac{300}{2000} = 1 - 0.15 = 0.85 = 85\%$$

Comparison:

$$\eta_{Otto} = 56.5\% < \eta_{Carnot} = 85\%$$

Otto is less efficient because:

• Uses isochoric (not isothermal) heat addition/rejection
• Doesn’t operate at constant max/min temperatures throughout
• Has irreversibilities

But 56.5% is still much better than real engines (~30%) due to friction, heat loss, etc.

Q6. A refrigerator (reverse Carnot engine) has COP = 5 and rejects 600 J to room. Find: (a) Heat removed from inside (b) Work input (c) If used as heat pump, what is COP?

Solution

(a) Heat removed:

$$\text{COP}_R = \frac{Q_C}{W}$$

Also: $Q_H = Q_C + W = 600$ J

$$\text{COP}_R = \frac{Q_C}{Q_H - Q_C}$$ $$5 = \frac{Q_C}{600 - Q_C}$$ $$5(600 - Q_C) = Q_C$$ $$3000 - 5Q_C = Q_C$$ $$3000 = 6Q_C$$ $$Q_C = 500 \text{ J}$$

(b) Work input:

$$W = Q_H - Q_C = 600 - 500 = 100 \text{ J}$$

Or: $W = Q_C/\text{COP}_R = 500/5 = 100$ J ✓

(c) COP as heat pump:

$$\text{COP}_{HP} = \text{COP}_R + 1 = 5 + 1 = 6$$

Or: $\text{COP}_{HP} = Q_H/W = 600/100 = 6$ ✓

Same device, different purpose → Different COP!

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Level 3: JEE Advanced

Q7. A Carnot engine operates between $T_H$ and $T_C$. If both temperatures are increased by same amount ΔT, does efficiency increase, decrease, or stay same? Prove mathematically.

Solution

Initial efficiency:

$$\eta_1 = 1 - \frac{T_C}{T_H}$$

After increasing both by ΔT:

$$\eta_2 = 1 - \frac{T_C + \Delta T}{T_H + \Delta T}$$

Compare:

$$\eta_2 - \eta_1 = \left(1 - \frac{T_C + \Delta T}{T_H + \Delta T}\right) - \left(1 - \frac{T_C}{T_H}\right)$$ $$= \frac{T_C}{T_H} - \frac{T_C + \Delta T}{T_H + \Delta T}$$ $$= \frac{T_C(T_H + \Delta T) - T_H(T_C + \Delta T)}{T_H(T_H + \Delta T)}$$ $$= \frac{T_C T_H + T_C\Delta T - T_H T_C - T_H\Delta T}{T_H(T_H + \Delta T)}$$ $$= \frac{T_C\Delta T - T_H\Delta T}{T_H(T_H + \Delta T)}$$ $$= \frac{\Delta T(T_C - T_H)}{T_H(T_H + \Delta T)}$$

Since $T_C < T_H$ (always), numerator is negative.

$$\eta_2 - \eta_1 < 0$$ $$\boxed{\eta_2 < \eta_1}$$

Efficiency DECREASES!

Intuition:

$$\eta = \frac{T_H - T_C}{T_H}$$

Adding same ΔT to both:

$$\eta_{new} = \frac{(T_H + \Delta T) - (T_C + \Delta T)}{T_H + \Delta T} = \frac{T_H - T_C}{T_H + \Delta T}$$

Same numerator, larger denominator → smaller efficiency.

Practical: To increase efficiency, increase $T_H$ MORE than $T_C$ (or decrease $T_C$).

Q8. A Carnot engine takes 1000 J from 500 K reservoir and rejects to 300 K reservoir. A second engine takes 1000 J from 300 K reservoir and rejects to 100 K reservoir. Compare: (a) Work outputs (b) Efficiencies (c) Heat rejected

Solution

Engine 1: 500 K to 300 K

$$\eta_1 = 1 - \frac{300}{500} = 0.4$$ $$W_1 = \eta_1 \times Q_{H1} = 0.4 \times 1000 = 400 \text{ J}$$ $$Q_{C1} = 1000 - 400 = 600 \text{ J}$$

Engine 2: 300 K to 100 K

$$\eta_2 = 1 - \frac{100}{300} = 1 - \frac{1}{3} = \frac{2}{3} ≈ 0.667$$ $$W_2 = \eta_2 \times Q_{H2} = \frac{2}{3} \times 1000 = 666.7 \text{ J}$$ $$Q_{C2} = 1000 - 666.7 = 333.3 \text{ J}$$

(a) Work comparison:

$$W_2 = 666.7 \text{ J} > W_1 = 400 \text{ J}$$

Engine 2 produces more work!

(b) Efficiency comparison:

$$\eta_2 = 66.7\% > \eta_1 = 40\%$$

Engine 2 is more efficient!

Why? Larger temperature ratio:

$$\frac{T_H - T_C}{T_H}\bigg|_1 = \frac{200}{500} = 0.4$$ $$\frac{T_H - T_C}{T_H}\bigg|_2 = \frac{200}{300} = 0.667$$

(c) Heat rejected:

$$Q_{C1} = 600 \text{ J} > Q_{C2} = 333.3 \text{ J}$$

Engine 1 wastes more heat.

Lesson: Even with same heat input, efficiency depends on temperature ratio, not absolute temperatures!

Q9. A Carnot refrigerator works between 0°C and 27°C. To make 10 kg of ice at 0°C from water at 0°C, how much work is required? (Latent heat of ice = 336 kJ/kg)

Solution

Given:

• $T_C = 0°C = 273$ K (ice compartment)
• $T_H = 27°C = 300$ K (room)
• Heat to remove: $Q_C = mL = 10 \times 336 = 3360$ kJ

COP of Carnot refrigerator:

$$\text{COP} = \frac{T_C}{T_H - T_C} = \frac{273}{300 - 273} = \frac{273}{27} = 10.11$$

Work required:

$$W = \frac{Q_C}{\text{COP}} = \frac{3360}{10.11} = 332.3 \text{ kJ}$$

Heat rejected to room:

$$Q_H = Q_C + W = 3360 + 332.3 = 3692.3 \text{ kJ}$$

Note: For every 1 kJ of work, refrigerator removes 10.11 kJ of heat (COP = 10.11)!

If refrigerator were 50% of Carnot efficiency:

$$\text{COP}_{real} = 0.5 \times 10.11 = 5.055$$ $$W_{real} = \frac{3360}{5.055} = 665 \text{ kJ}$$

(twice as much!)

Q10. Derive the efficiency of Otto cycle in terms of compression ratio r and γ.

Solution

Otto cycle processes:

1. 1→2: Adiabatic compression, $V_1$ to $V_2$
2. 2→3: Isochoric heat addition at $V_2$
3. 3→4: Adiabatic expansion, $V_2$ to $V_1$
4. 4→1: Isochoric heat rejection at $V_1$

Compression ratio: $r = V_1/V_2 = V_{max}/V_{min}$

Heat addition (isochoric 2→3):

$$Q_{in} = nC_V(T_3 - T_2)$$

Heat rejection (isochoric 4→1):

$$Q_{out} = nC_V(T_4 - T_1)$$

Efficiency:

$$\eta = 1 - \frac{Q_{out}}{Q_{in}} = 1 - \frac{T_4 - T_1}{T_3 - T_2}$$

For adiabatic processes:

1→2: $T_1V_1^{\gamma-1} = T_2V_2^{\gamma-1}$

$$\frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{\gamma-1} = r^{\gamma-1}$$

3→4: $T_3V_2^{\gamma-1} = T_4V_1^{\gamma-1}$

$$\frac{T_3}{T_4} = \left(\frac{V_1}{V_2}\right)^{\gamma-1} = r^{\gamma-1}$$

Therefore:

$$\frac{T_2}{T_1} = \frac{T_3}{T_4} = r^{\gamma-1}$$

Or: $T_2 = T_1 r^{\gamma-1}$ and $T_3 = T_4 r^{\gamma-1}$

Substitute in efficiency:

$$\eta = 1 - \frac{T_4 - T_1}{T_3 - T_2} = 1 - \frac{T_4 - T_1}{T_4 r^{\gamma-1} - T_1 r^{\gamma-1}}$$ $$= 1 - \frac{T_4 - T_1}{r^{\gamma-1}(T_4 - T_1)}$$

$$\boxed{\eta_{Otto} = 1 - \frac{1}{r^{\gamma-1}}}$$

Depends only on compression ratio and γ!

To increase efficiency: Increase r (higher compression)

Typical values:

• r = 8, γ = 1.4 → η ≈ 56%
• r = 10, γ = 1.4 → η ≈ 60%

Real engines: ~30% (half of theoretical due to irreversibilities)

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Connection to Other Topics

→ Second Law of Thermodynamics

Carnot theorem, efficiency limits: Second Law →

→ PV Diagrams

Carnot cycle, Otto cycle on PV diagrams: PV Diagrams →

→ Entropy

Carnot cycle has zero entropy change: Entropy →

→ Thermodynamic Processes

Isothermal, adiabatic processes in cycles: Processes →

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Quick Revision Formula Sheet

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JEE Strategy Tips

1. Always convert to Kelvin for efficiency formulas
2. Carnot is maximum - any real engine is less efficient
3. Series engines: Overall η = Carnot from $T_{highest}$ to $T_{lowest}$
4. Compression ratio: Higher r → Higher Otto efficiency
5. COP can be > 1 (refrigerator/heat pump), but η < 1 (engine)
6. Remember: $\text{COP}_{HP} = \text{COP}_R + 1$
7. For Carnot: Only temperatures matter, not working substance

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Next Topic: Entropy →

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Last updated: February 2025