Q.33 Physics ⭐ 360° Option Elimination Active

Vault ↗

NEET 2023 Physics Question 33 Solution

Topic: Thermodynamics · Official NTA Booklet Code G5

Question

A Carnot engine has an efficiency of 50% when its source is at a temperature 327°C. The temperature of the sink is:

✓

Verified Against NTA Final Answer Key

Option (2) : 27° C

📚 Thermodynamics Easy

📋 Given Parameters & Problem Setup

Carnot efficiency (η) 50% = 0.5

Source temperature (T₁) 327° C = 327 + 273 = 600 K

Sink temperature (T₂) To be determined in °C

⚡

The 30-Second Exam Hall Shortcut: ⚡ Carnot Efficiency Rule: T₂ = T₁ (1 - η) in Kelvin

⏱️ Save ~40 seconds

Convert to Kelvin: T₁ = 327 + 273 = 600 K. With 50% efficiency, T₂ = 600 × 0.5 = 300 K. Convert back to Celsius: 300 - 273 = 27° C in 15 seconds!

1 ALWAYS convert Celsius to absolute Kelvin scale first: T₁ = 327 + 273 = 600 K.

2 Efficiency η = 1 - (T₂ / T₁) = 0.5 ⟹ T₂ / T₁ = 0.5.

3 Solve for sink temperature in Kelvin: T₂ = 0.5 × 600 K = 300 K.

4 Convert back to Celsius: T₂ = 300 - 273 = 27° C.

5 Option (2) 27° C is locked instantly!

🛡️

Why You Can 100% Trust This Shortcut: Thermodynamics · Carnot Theorem (Carnot cycle efficiency depends strictly on absolute thermodynamic temperatures: η = 1 - T_sink / T_source).

📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Conversion of Source Temperature to Kelvin

Thermodynamic efficiency formulas are strictly valid only on the absolute temperature scale (Kelvin). Convert the source temperature T₁ from Celsius to Kelvin:

T_1 = 327^\circ\text{C} + 273 = 600\text{ K}

Step 2

Governing Formula for Carnot Engine Efficiency

The efficiency η of a reversible Carnot engine operating between a source at temperature T₁ and a sink at temperature T₂ is given by:

\eta = 1 - \frac{T_2}{T_1}

Step 3

Substitution and Calculation of Sink Temperature in Kelvin

Given efficiency η = 50% = 0.5 and T₁ = 600 K:

0.5 = 1 - \frac{T_2}{600} \implies \frac{T_2}{600} = 0.5 \implies T_2 = 300\text{ K}

Step 4

Conversion of Sink Temperature to Celsius

Convert T₂ back into degrees Celsius:

T_2(^{\circ}\text{C}) = 300\text{ K} - 273 = 27^\circ\text{C}

⭐

360° Option Elimination Engine Key Feature

Diagnose every choice: Know exactly why the right option works and why distractors fail.

4-Option Deep Scan

1 (1)

✗ Incorrect

Verdict & Scientific Reason:

Arbitrary wrong value.

⚠️

Student Fallacy (Why candidates pick this trap): Candidate subtracted 100 from 300.

2 (2)

✓ 100% Correct

Verdict & Scientific Reason:

Accurate sink temperature: 300 K = 27° C.

💡

Why this is airtight: None. Candidate converted to Kelvin before applying efficiency formula.

3 (3)

✗ Incorrect

Verdict & Scientific Reason:

Incorrect temperature calculation.

⚠️

Student Fallacy (Why candidates pick this trap): Candidate made an arithmetic blunder.

4 (4)

✗ Incorrect

Verdict & Scientific Reason:

Candidate used Celsius directly (327 / 2 = 163.5° C) or made a standard temperature error.

⚠️

Student Fallacy (Why candidates pick this trap): Candidate forgot the fundamental Kelvin conversion.

⚠️

NTA Trap Alert: Using Temperatures in Celsius Directly Without Converting to Kelvin

If a student calculates directly in Celsius: T₂ = 327 × (1 - 0.5) = 163.5° C. The Carnot formula is DERIVED from the Kelvin scale (dQ/T = 0). Using Celsius directly is a severe conceptual violation.

🛡️ How to avoid the -1 mark: Rule of Thumb in Thermodynamics: ALWAYS add 273 first, solve in Kelvin, then subtract 273.

📖

Core NEET Syllabus Standard

National Eligibility cum Entrance Test (NEET UG) · Thermodynamics Standard Benchmark

Core Scientific Principle Tested:

"Carnot efficiency is η = 1 - T_sink / T_source, where temperatures must be in Kelvin."

Syllabus Focus: Carnot Engine and Efficiency

Preserved in Rationalized Curriculum Syllabus & Academic Standards ↗

Ready to attempt under real exam pressure?

Test your speed and accuracy in the authentic 200-minute timed NTA simulator.

Launch NEET 2023 CBT Test →

← Previous (Q.32) All Physics Questions Grid Next (Q.34) →

Equipartition of Energy: Determining Vibrational Modes f from Specific Heat Ratio C_P / C_V

View Solution →

NEET 2026 Q59

Entropy Changes in Isothermal Free Expansion of an Ideal Gas

View Solution →

Spotted a mistake in this solution? Report an error (support@mycompetitiveexams.com)

⚖️ Legal & Fair Dealing Notice: Sourced from publicly released NEET 2023 question papers and NEET answer keys for private study, research, and non-commercial educational practice under Fair Dealing (Section 52(1)(a) of the Indian Copyright Act, 1957). Not affiliated with or endorsed by NTA or NMC.

NEET 2023 Physics Question 33 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Carnot Efficiency Rule: T₂ = T₁ (1 - η) in Kelvin

Step-by-Step Rigorous Workflow & Mathematical Proof

Conversion of Source Temperature to Kelvin

Governing Formula for Carnot Engine Efficiency

Substitution and Calculation of Sink Temperature in Kelvin

Conversion of Sink Temperature to Celsius

360° Option Elimination Engine Key Feature

NTA Trap Alert: Using Temperatures in Celsius Directly Without Converting to Kelvin

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

🔗 Related Questions — Thermodynamics

Work Done by an Ideal Monatomic Gas in Adiabatic Expansion

Net Heat Supplied in a Cyclic PV Indicator Diagram

Equipartition of Energy: Determining Vibrational Modes f from Specific Heat Ratio C_P / C_V

Entropy Changes in Isothermal Free Expansion of an Ideal Gas

Solve 4 NEET Questions a Day — One per Subject