Physics Fluids and Thermal - How to Solve: Carnot Engine & Refrigerator (Efficiency, COP) – IIT JEE Guide

How to Solve: Carnot Engine & Refrigerator (Efficiency, COP) – IIT JEE Guide

(For Students & Teachers – Ready-to-Record Script Included)

Introduction

Mastering Carnot engines and refrigerators unlocks 5-8 marks in IIT JEE—enough to push you from a 90 to a 99+ percentile. These questions test your ability to apply thermodynamics to real-world machines, and examiners love disguising them in word problems. If you follow this method, you’ll solve them in under 2 minutes—even under exam pressure.

WHAT YOU NEED TO KNOW FIRST

Before diving in, ensure you understand: 1. First Law of Thermodynamics (ΔU = Q – W) – Energy conservation in heat engines. 2. Isothermal & Adiabatic Processes – How temperature and heat transfer work in reversible cycles. 3. Kelvin & Celsius Temperature Scales – You’ll convert between them often.

(If any of these are shaky, pause and review them first.)

KEY TERMS & FORMULAS

Key Terms

1. Carnot Engine – An ideal, reversible heat engine with maximum possible efficiency.
2. Refrigerator (Heat Pump) – A device that transfers heat from a cold reservoir to a hot reservoir using work.
3. Efficiency (η) – Fraction of heat input converted to work. Higher η = better engine.
4. Coefficient of Performance (COP) – Measure of a refrigerator’s effectiveness. Higher COP = better fridge.

Formulas

1. Carnot Engine Efficiency (η)

Formula: [ \eta = 1 - \frac{T_C}{T_H} ] - (T_H) = Temperature of hot reservoir (in Kelvin) - (T_C) = Temperature of cold reservoir (in Kelvin) - MEMORISE THIS – It’s the maximum possible efficiency for any heat engine.

Alternative Form (if given heat instead of temperature): [ \eta = 1 - \frac{Q_C}{Q_H} ] - (Q_H) = Heat absorbed from hot reservoir - (Q_C) = Heat rejected to cold reservoir - Given on exam sheet (but memorising saves time).

2. Refrigerator COP (COP_ref)

Formula: [ COP_{ref} = \frac{T_C}{T_H - T_C} ] - (T_C) = Temperature of cold reservoir (inside fridge) - (T_H) = Temperature of hot reservoir (outside air) - MEMORISE THIS – It’s the maximum possible COP for any refrigerator.

Alternative Form (if given heat/work): [ COP_{ref} = \frac{Q_C}{W} ] - (Q_C) = Heat removed from cold reservoir - (W) = Work input (electricity used) - Given on exam sheet (but useful for problem-solving).

3. Heat Pump COP (COP_hp)

Formula: [ COP_{hp} = \frac{T_H}{T_H - T_C} ] - (T_H) = Temperature of hot reservoir (inside room) - (T_C) = Temperature of cold reservoir (outside air) - MEMORISE THIS – Used when the device heats a space (e.g., a room heater).

Alternative Form: [ COP_{hp} = \frac{Q_H}{W} ] - (Q_H) = Heat delivered to hot reservoir - (W) = Work input - Given on exam sheet.

4. Relationship Between COP_ref and COP_hp

[ COP_{hp} = COP_{ref} + 1 ] - MEMORISE THIS – Useful for quick conversions between fridge and heat pump modes.

STEP-BY-STEP METHOD

For Carnot Engine Problems (Efficiency)

Step 1: Identify (T_H) and (T_C) from the problem. - If given in Celsius, convert to Kelvin (add 273). - Example: (T_H = 500°C = 773 K), (T_C = 27°C = 300 K).

Step 2: Plug into the efficiency formula: [ \eta = 1 - \frac{T_C}{T_H} ]

Step 3: Calculate and express as a percentage (multiply by 100 if needed). - Example: (\eta = 1 - \frac{300}{773} = 0.612 \rightarrow 61.2\%).

Step 4: If asked for work output ((W)), use: [ W = \eta \times Q_H ]

For Refrigerator/Heat Pump Problems (COP)

Step 1: Determine if it’s a refrigerator (cooling) or heat pump (heating). - Refrigerator: Removes heat from cold space → Use (COP_{ref}). - Heat Pump: Adds heat to hot space → Use (COP_{hp}).

Step 2: Identify (T_H) and (T_C) (in Kelvin). - Example: Fridge at (5°C (278 K)), room at (25°C (298 K)).

Step 3: Plug into the correct COP formula: - For fridge: (COP_{ref} = \frac{T_C}{T_H - T_C}) - For heat pump: (COP_{hp} = \frac{T_H}{T_H - T_C})

Step 4: If given (Q_C) or (Q_H), use the alternative forms: - (COP_{ref} = \frac{Q_C}{W}) - (COP_{hp} = \frac{Q_H}{W})

Step 5: Solve for the unknown (usually (W) or (Q)).

WORKED EXAMPLES

Example 1 – Basic Carnot Engine

Problem: A Carnot engine operates between (300 K) and (600 K). What is its efficiency?

Solution: Step 1: Identify (T_H = 600 K), (T_C = 300 K). Step 2: Plug into efficiency formula: [ \eta = 1 - \frac{T_C}{T_H} = 1 - \frac{300}{600} = 1 - 0.5 = 0.5 ] Step 3: Convert to percentage: (0.5 \times 100 = 50\%).

What we did and why: We used the Carnot efficiency formula because the problem gave temperatures. The key was converting to Kelvin (though here it was already given). Efficiency is always ≤ 100%, so 50% is reasonable.

Example 2 – Medium Refrigerator COP

Problem: A refrigerator maintains (5°C) inside while the room is (27°C). If it removes (1200 J) of heat from the inside, how much work is done?

Solution: Step 1: Convert temperatures to Kelvin: - (T_C = 5°C = 278 K) - (T_H = 27°C = 300 K)

Step 2: Calculate (COP_{ref}): [ COP_{ref} = \frac{T_C}{T_H - T_C} = \frac{278}{300 - 278} = \frac{278}{22} \approx 12.64 ]

Step 3: Use (COP_{ref} = \frac{Q_C}{W}) to find (W): [ 12.64 = \frac{1200}{W} \implies W = \frac{1200}{12.64} \approx 95 J ]

What we did and why: We converted Celsius to Kelvin first (critical step!). Then, we used the COP formula to relate heat removed ((Q_C)) to work input ((W)). The high COP (12.64) means the fridge is efficient—it removes 12.64 J of heat per 1 J of work.

Example 3 – Exam-Style (Disguised Problem)

Problem: An ideal heat engine operates between two reservoirs at (227°C) and (27°C). If it absorbs (1000 J) of heat from the hot reservoir, what is the work done?

Solution: Step 1: Convert temperatures to Kelvin: - (T_H = 227°C = 500 K) - (T_C = 27°C = 300 K)

Step 2: Calculate efficiency: [ \eta = 1 - \frac{T_C}{T_H} = 1 - \frac{300}{500} = 0.4 \quad (40\%) ]

Step 3: Use (W = \eta \times Q_H): [ W = 0.4 \times 1000 = 400 J ]

What we did and why: The problem disguised the Carnot engine as a "heat engine." We converted temperatures first, then used efficiency to find work. The key was recognising it’s a Carnot problem despite the wording.

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP (Night Before Exam)

"Listen up—this is your 60-second Carnot crash course. First, efficiency for a Carnot engine is (1 - \frac{T_C}{T_H}), always in Kelvin. If they give you heat, use (1 - \frac{Q_C}{Q_H}). For refrigerators, COP is (\frac{T_C}{T_H - T_C}), and for heat pumps, it’s (\frac{T_H}{T_H - T_C}). Remember: Kelvin only, hot reservoir is always the bigger number, and COP can be >1 (that’s normal!). If they ask for work, use (W = \eta Q_H) for engines or (W = \frac{Q_C}{COP}) for fridges. Watch out for Celsius traps—convert first! Now go crush those 5-8 marks."