Real-Life Hook: Why Can’t We Run Cars on Ocean Heat?
The ocean contains enormous thermal energy—enough to power civilization for millennia! So why can’t we extract it and run cars, factories, and cities for free?
The answer lies in the Second Law of Thermodynamics: you can’t convert heat completely into work without something else happening. The ocean is at uniform temperature, and the Second Law forbids extracting work from a single heat reservoir.
This explains why:
- We can’t build perpetual motion machines
- Refrigerators need external power
- All real engines are less than 100% efficient
- Time flows in one direction (broken eggs don’t un-break!)
The Second Law is the most profound law in physics—it governs the arrow of time itself.
The Second Law: Why It Matters
What Does the First Law Miss?
First Law: Energy is conserved (Q = ΔU + W)
But First Law allows:
- Heat flowing from cold to hot spontaneously ❌
- 100% conversion of heat to work ❌
- Running engines on single reservoir ❌
None of these happen in nature! We need the Second Law.
The Second Law: Multiple Perspectives
The Second Law has several equivalent statements:
- Kelvin-Planck Statement (heat engines)
- Clausius Statement (refrigerators)
- Entropy Statement (general)
All are equivalent—proving one proves all!
Kelvin-Planck Statement
Statement
It is impossible to construct a heat engine that, operating in a cycle, produces no effect other than the absorption of heat from a reservoir and the performance of an equal amount of work.
Simplified Version
You cannot build an engine that:
- Takes heat from single source
- Converts it completely to work (100% efficiency)
- Has no other effect
Why It Matters
Implications:
- No perfect heat engine (efficiency always < 100%)
- Must have two reservoirs (hot and cold)
- Must reject some heat to cold reservoir
- Carnot engine is the best possible
Violation Would Mean
If Kelvin-Planck could be violated:
- Ships could run on ocean heat alone
- Cars could run on atmospheric heat
- Perpetual motion of second kind
- Free energy from environment!
Nature forbids this. Some heat MUST be wasted.
Mathematical Form
For any heat engine:
$$\eta = \frac{W}{Q_H} < 1$$Or equivalently:
$$W < Q_H$$Cannot convert all absorbed heat to work in a cycle.
Clausius Statement
Statement
It is impossible to construct a device that, operating in a cycle, produces no effect other than the transfer of heat from a cooler body to a hotter body.
Simplified Version
Heat cannot spontaneously flow from cold to hot.
Why It Matters
Implications:
- Heat flows cold → hot only with external work
- Refrigerators need power input
- Air conditioners require electricity
- Natural direction: hot → cold
Violation Would Mean
If Clausius could be violated:
- Rooms would cool without AC
- Ice would form in hot water spontaneously
- Refrigerators would work without power
- Heat would flow “uphill” for free
Never observed in nature.
Mathematical Form
For refrigerator/heat pump:
$$\text{Work input } W > 0 \text{ required to move heat from } T_C \text{ to } T_H$$Equivalence of Kelvin-Planck and Clausius
Proof: Clausius Violation → Kelvin-Planck Violation
Assume: We have device R violating Clausius (transfers Q from cold to hot with no work).
Construct:
- Device R: Transfers Q₂ from cold to hot (no work)
- Real engine E: Takes Q₁ from hot, does work W, rejects Q₂ to cold
Combined system:
- Hot reservoir: receives Q₂ from R, gives Q₁ to E → net: receives (Q₂ - Q₁) = -W
- Cold reservoir: gives Q₂ to R, receives Q₂ from E → net: zero
- Output: Work W
Result: Combined device takes heat W from single reservoir (hot) and converts to work W completely!
This violates Kelvin-Planck!
Therefore: Clausius violation → Kelvin-Planck violation
Proof: Kelvin-Planck Violation → Clausius Violation
Left as exercise (similar logic, reverse direction)
Conclusion: The two statements are equivalent.
Heat Engine Fundamentals
Heat Engine Basics
Components:
- Hot reservoir at $T_H$: Supplies heat $Q_H$
- Working substance (gas): Undergoes cyclic process
- Cold reservoir at $T_C$: Receives rejected heat $Q_C$
Process:
- Absorb heat $Q_H$ from hot reservoir
- Do work $W$ on surroundings
- Reject heat $Q_C$ to cold reservoir
- Return to initial state (cyclic)
Energy Conservation (First Law)
For complete cycle: $\Delta U = 0$
$$Q_H = W + Q_C$$ $$W = Q_H - Q_C$$Efficiency
Efficiency:
$$\eta = \frac{W}{Q_H} = \frac{Q_H - Q_C}{Q_H}$$ $$\eta = 1 - \frac{Q_C}{Q_H}$$Second Law constraint:
$$\eta < 1 \quad \text{(always)}$$Maximum Efficiency: Carnot Engine
Carnot efficiency:
$$\eta_{Carnot} = 1 - \frac{T_C}{T_H}$$This is the maximum possible efficiency for any engine between $T_H$ and $T_C$.
Key point: Depends ONLY on temperatures, not on working substance!
More details: Heat Engines and Carnot Cycle →
Refrigerator and Heat Pump
Refrigerator
Purpose: Remove heat from cold space
Components:
- Cold reservoir at $T_C$ (inside fridge)
- Hot reservoir at $T_H$ (room)
- Work input $W$
Process:
- Absorb heat $Q_C$ from cold space
- Input work $W$
- Reject heat $Q_H$ to room
- $Q_H = Q_C + W$
Coefficient of Performance (COP)
For Refrigerator:
$$\text{COP}_R = \frac{Q_C}{W} = \frac{Q_C}{Q_H - Q_C}$$Higher is better (more cooling per unit work)
Carnot refrigerator (maximum):
$$\text{COP}_{Carnot,R} = \frac{T_C}{T_H - T_C}$$Heat Pump
Purpose: Heat warm space (opposite focus from refrigerator)
For Heat Pump:
$$\text{COP}_{HP} = \frac{Q_H}{W} = \frac{Q_H}{Q_H - Q_C}$$Note: $\text{COP}_{HP} = \text{COP}_R + 1$
Carnot heat pump:
$$\text{COP}_{Carnot,HP} = \frac{T_H}{T_H - T_C}$$Key Differences
| Device | Goal | COP | Typical Values |
|---|---|---|---|
| Refrigerator | Cool inside | $Q_C/W$ | 2-5 |
| Heat Pump | Heat inside | $Q_H/W$ | 3-6 |
| Heat Engine | Produce work | $W/Q_H$ (η) | 0.3-0.6 |
Note: COP can be > 1 (unlike efficiency)! This is because refrigerator moves existing heat, doesn’t create it.
Reversible vs Irreversible Processes
Reversible Process
Definition: A process that can be reversed by infinitesimal changes, leaving no trace in universe.
Conditions:
- Quasi-static: Infinitely slow (always near equilibrium)
- No friction: No dissipative forces
- No heat flow across finite ΔT: Only across infinitesimal temperature difference
- No turbulence, mixing, etc.
Examples:
- Ideal gas expanding in frictionless piston, infinitely slowly
- Carnot cycle (theoretical)
Reality: No real process is perfectly reversible!
Irreversible Process
Definition: Process that cannot be reversed without leaving changes in surroundings.
Causes of Irreversibility:
- Friction
- Heat transfer across finite ΔT
- Free expansion
- Mixing of different substances
- Plastic deformation
- Electrical resistance
Examples:
- Rubbing hands (friction → heat)
- Ice melting in warm water
- Gas expanding into vacuum
- Stirring cream into coffee
Reality: All real processes are irreversible!
Why It Matters
Key Theorem:
All reversible engines operating between same two temperatures have same efficiency (Carnot efficiency). All irreversible engines have lower efficiency.
Carnot engine is the ideal limit!
Carnot’s Theorem
Statement
No engine operating between two given temperatures can be more efficient than a Carnot (reversible) engine operating between the same temperatures.
Implications
- Carnot efficiency is maximum possible
- All reversible engines have same efficiency (independent of working substance!)
- All real engines are less efficient (due to irreversibilities)
- Efficiency depends only on temperatures
Proof (by Contradiction)
Assume: Engine E is more efficient than Carnot engine C.
Setup:
- Both operate between $T_H$ and $T_C$
- Let $\eta_E > \eta_C$
- Run C in reverse as refrigerator (possible since reversible)
Energy flow:
- Engine E: Takes $Q_H$ from hot, does work $W_E$, rejects $Q_C$
- Refrigerator C: Takes work $W_C < W_E$, removes $Q_C'$ from cold, delivers $Q_H'$ to hot
Adjust C so $Q_H' = Q_H$:
- Work input to C: $W_C = Q_H(1 - \eta_C)$
- Work output from E: $W_E = Q_H \eta_E$
- Net work: $W_{net} = W_E - W_C = Q_H(\eta_E - \eta_C) > 0$
Combined system:
- Hot reservoir: Zero net heat transfer
- Cold reservoir: Net heat extracted
- Output: Net work
This is a device extracting heat from single reservoir and converting completely to work!
Violates Kelvin-Planck!
Therefore: $\eta_E \leq \eta_{Carnot}$
Common Mistakes & Misconceptions
Mistake 1: Confusing Efficiency and COP
❌ Refrigerator efficiency = $Q_C/W$ ✅ Refrigerator COP = $Q_C/W$ (don’t call it efficiency!)
Efficiency applies to engines (always < 1). COP applies to refrigerators/heat pumps (can be > 1).
Mistake 2: 100% Efficiency Possible
❌ With perfect insulation, η = 100% ✅ Second Law forbids η = 100% (even with perfect insulation)
Must reject some heat to cold reservoir.
Mistake 3: Second Law = Energy Conservation
❌ Second Law says energy is conserved ✅ That’s First Law! Second Law says direction of processes
First Law: How much energy? Second Law: Which direction?
Mistake 4: Reversible = Can Physically Reverse
❌ Reversible means you can run it backward in practice ✅ Reversible means thermodynamically reversible (quasi-static, no friction, etc.)
Real processes can often run backward (like engines), but they’re still thermodynamically irreversible!
Memory Trick: “KP-HE, C-R”
- Kelvin-Planck: Heat Engines (can’t be 100% efficient)
- Clausius: Refrigerators (need work to cool)
Trick: Carnot is the GOAT
Greatest Of All Temperatures (efficiency depends only on T)
- Best possible efficiency
- Same for all working substances
- Only temperature matters
Practice Problems
Level 1: JEE Main Basics
Q1. Which statement is the Kelvin-Planck statement of Second Law? (a) Heat cannot flow from cold to hot (b) No engine can be 100% efficient (c) Entropy always increases (d) Energy is conserved
Solution
(b) No engine can be 100% efficient
More precisely: No engine operating in a cycle can convert heat completely to work without any other effect.
- (a) is Clausius statement
- (c) is Entropy statement
- (d) is First Law
Q2. A Carnot engine operates between 500 K and 300 K. Find its efficiency.
Solution
$$\eta = 1 - \frac{T_C}{T_H} = 1 - \frac{300}{500} = 1 - 0.6 = 0.4$$Efficiency = 40%
Remember: Always use Kelvin for Carnot efficiency!
Q3. A refrigerator absorbs 100 J from cold reservoir and rejects 140 J to hot reservoir. Find: (a) Work input (b) COP
Solution
(a) Work input:
$$Q_H = Q_C + W$$ $$140 = 100 + W$$ $$W = 40 \text{ J}$$(b) COP:
$$\text{COP} = \frac{Q_C}{W} = \frac{100}{40} = 2.5$$Level 2: JEE Main/Advanced
Q4. A heat engine has efficiency 30%. It rejects 700 J to cold reservoir per cycle. Find: (a) Heat absorbed from hot reservoir (b) Work done per cycle
Solution
Given: $\eta = 0.3$, $Q_C = 700$ J
(a) Heat absorbed:
$$\eta = 1 - \frac{Q_C}{Q_H}$$ $$0.3 = 1 - \frac{700}{Q_H}$$ $$\frac{700}{Q_H} = 0.7$$ $$Q_H = \frac{700}{0.7} = 1000 \text{ J}$$(b) Work done:
$$W = Q_H - Q_C = 1000 - 700 = 300 \text{ J}$$Or: $W = \eta \times Q_H = 0.3 \times 1000 = 300$ J ✓
Q5. An ideal refrigerator operates between -10°C and 30°C. Find: (a) Maximum COP (b) Heat rejected to room when 100 J is removed from cold space
Solution
Convert to Kelvin: $T_C = 263$ K, $T_H = 303$ K
(a) Maximum COP (Carnot):
$$\text{COP}_{max} = \frac{T_C}{T_H - T_C} = \frac{263}{303 - 263} = \frac{263}{40} = 6.575$$(b) Work required:
$$\text{COP} = \frac{Q_C}{W}$$ $$W = \frac{Q_C}{\text{COP}} = \frac{100}{6.575} = 15.21 \text{ J}$$Heat rejected:
$$Q_H = Q_C + W = 100 + 15.21 = 115.21 \text{ J}$$Note: High COP means little work needed (temperature difference is small).
Q6. Two Carnot engines A and B operate in series. Engine A absorbs heat from reservoir at 600 K and rejects to reservoir at temperature T. Engine B absorbs this rejected heat and rejects to reservoir at 300 K. If both have same efficiency, find T.
Solution
Engine A: Between 600 K and T
$$\eta_A = 1 - \frac{T}{600}$$Engine B: Between T and 300 K
$$\eta_B = 1 - \frac{300}{T}$$Given: $\eta_A = \eta_B$
$$1 - \frac{T}{600} = 1 - \frac{300}{T}$$ $$\frac{T}{600} = \frac{300}{T}$$ $$T^2 = 600 \times 300 = 180000$$ $$T = \sqrt{180000} = 424.26 \text{ K}$$Or use geometric mean: For equal efficiencies in series, intermediate temperature is geometric mean:
$$T = \sqrt{T_H \times T_C} = \sqrt{600 \times 300} = \sqrt{180000} ≈ 424 \text{ K}$$Level 3: JEE Advanced
Q7. A heat engine absorbs 1000 J from reservoir at 500 K and rejects heat to reservoir at 300 K. What is the maximum work it can do? If it does only 300 J of work, is the engine reversible?
Solution
(a) Maximum work (Carnot):
$$\eta_{max} = 1 - \frac{300}{500} = 1 - 0.6 = 0.4$$ $$W_{max} = \eta_{max} \times Q_H = 0.4 \times 1000 = 400 \text{ J}$$(b) Is engine reversible?
Actual work: $W = 300$ J < $W_{max} = 400$ J
Actual efficiency:
$$\eta_{actual} = \frac{300}{1000} = 0.3 < \eta_{max}$$Engine is irreversible (less efficient than Carnot).
Heat rejected:
$$Q_C = Q_H - W = 1000 - 300 = 700 \text{ J}$$Check: For Carnot, $Q_C$ would be $1000 - 400 = 600$ J. Actual engine rejects more heat (700 J > 600 J), confirming irreversibility.
Q8. Prove that COP of heat pump is always greater than COP of refrigerator by exactly 1.
Solution
For Refrigerator:
$$\text{COP}_R = \frac{Q_C}{W}$$For Heat Pump:
$$\text{COP}_{HP} = \frac{Q_H}{W}$$From First Law: $Q_H = Q_C + W$
Substitute in $\text{COP}_{HP}$:
$$\text{COP}_{HP} = \frac{Q_C + W}{W} = \frac{Q_C}{W} + \frac{W}{W}$$ $$\text{COP}_{HP} = \frac{Q_C}{W} + 1$$ $$\text{COP}_{HP} = \text{COP}_R + 1$$Proved!
Intuition: Heat pump delivers all the heat removed from cold space PLUS the work done, so COP is higher by exactly the work term.
Example: If $\text{COP}_R = 4$, then $\text{COP}_{HP} = 5$.
Q9. A Carnot engine efficiency is 40%. If temperature of cold reservoir is decreased by 50 K, efficiency becomes 50%. Find the temperatures of both reservoirs.
Solution
Initial condition:
$$\eta_1 = 1 - \frac{T_C}{T_H} = 0.4$$ $$\frac{T_C}{T_H} = 0.6$$ $$T_C = 0.6T_H \quad ...(1)$$After decreasing $T_C$ by 50 K:
$$\eta_2 = 1 - \frac{T_C - 50}{T_H} = 0.5$$ $$\frac{T_C - 50}{T_H} = 0.5$$ $$T_C - 50 = 0.5T_H$$ $$T_C = 0.5T_H + 50 \quad ...(2)$$Equate (1) and (2):
$$0.6T_H = 0.5T_H + 50$$ $$0.1T_H = 50$$ $$T_H = 500 \text{ K}$$From (1):
$$T_C = 0.6 \times 500 = 300 \text{ K}$$Verification:
- Initial: $\eta = 1 - 300/500 = 0.4$ ✓
- After: $\eta = 1 - 250/500 = 0.5$ ✓
Answer: $T_H = 500$ K (227°C), $T_C = 300$ K (27°C)
Q10. Three Carnot engines operate in series between 800 K and 200 K with equal efficiencies. Find the intermediate temperatures.
Solution
Let temperatures be: 800 K, $T_1$, $T_2$, 200 K
Engines:
- A: 800 K to $T_1$, efficiency $\eta$
- B: $T_1$ to $T_2$, efficiency $\eta$
- C: $T_2$ to 200 K, efficiency $\eta$
For equal efficiencies:
$$1 - \frac{T_1}{800} = 1 - \frac{T_2}{T_1} = 1 - \frac{200}{T_2}$$From first two:
$$\frac{T_1}{800} = \frac{T_2}{T_1}$$ $$T_1^2 = 800T_2 \quad ...(1)$$From last two:
$$\frac{T_2}{T_1} = \frac{200}{T_2}$$ $$T_2^2 = 200T_1 \quad ...(2)$$Substitute (2) into (1):
$$T_1^2 = 800T_2$$From (2): $T_2 = \sqrt{200T_1}$
$$T_1^2 = 800\sqrt{200T_1}$$ $$T_1^4 = 640000 \times 200T_1$$ $$T_1^3 = 128000000$$ $$T_1 = (128 \times 10^6)^{1/3} = 502.4 \text{ K}$$Wait, let me use a better approach.
For n engines in series with equal efficiencies: Temperatures form geometric progression!
$$T_1 = 800 \times r$$ $$T_2 = 800 \times r^2$$ $$200 = 800 \times r^3$$ $$r^3 = \frac{200}{800} = \frac{1}{4}$$ $$r = \left(\frac{1}{4}\right)^{1/3} = \frac{1}{4^{1/3}} = \frac{1}{1.587} = 0.630$$ $$T_1 = 800 \times 0.630 = 504 \text{ K}$$ $$T_2 = 800 \times (0.630)^2 = 800 \times 0.397 = 317.6 \text{ K}$$Verification:
- $\eta_A = 1 - 504/800 = 0.37$
- $\eta_B = 1 - 317.6/504 = 0.37$ ✓
- $\eta_C = 1 - 200/317.6 = 0.37$ ✓
Answer: $T_1 ≈ 504$ K, $T_2 ≈ 318$ K
General formula: For n engines between $T_H$ and $T_C$:
$$T_i = T_H \left(\frac{T_C}{T_H}\right)^{i/n}$$Connection to Other Topics
→ Heat Engines and Carnot Cycle
Detailed analysis of Carnot engine, efficiency: Heat Engines →
→ Entropy
Second Law can be stated as “Entropy of universe increases”: Entropy →
→ PV Diagrams
Reversible processes are quasi-static paths on PV diagrams: PV Diagrams →
→ Real Engines
Otto, Diesel cycles less efficient than Carnot: Real engine cycles →
Quick Revision Points
Kelvin-Planck:
No engine can be 100% efficient in converting heat to work
Clausius:
Heat cannot spontaneously flow from cold to hot
Carnot Efficiency:
$$\eta_{Carnot} = 1 - \frac{T_C}{T_H}$$(Maximum possible between $T_H$ and $T_C$)
Refrigerator COP:
$$\text{COP}_R = \frac{Q_C}{W} = \frac{T_C}{T_H - T_C}$$Heat Pump COP:
$$\text{COP}_{HP} = \frac{Q_H}{W} = \frac{T_H}{T_H - T_C} = \text{COP}_R + 1$$Key Relations:
- All reversible engines: Same efficiency (Carnot)
- All irreversible engines: Lower efficiency
- Real processes: Always irreversible
JEE Strategy Tips
- Always convert to Kelvin for Carnot formulas
- Efficiency < 1 for engines, COP can be > 1 for refrigerators
- Carnot is maximum - any real engine has lower efficiency
- Reversible ≠ Realistic - all real processes are irreversible
- For series engines: Use geometric mean for equal efficiencies
- Remember: $\text{COP}_{HP} = \text{COP}_R + 1$
Next Topic: Heat Engines and Carnot Cycle →
Last updated: February 2025