Heat Engine : Second Law of Thermodynamics — Important Questions

50%

Step 1: Given: T₁ = 500 K (source temperature), T₂ = 250 K (sink temperature). Step 2: Using the Carnot efficiency formula: η = 1 - T₂/T₁. Step 3: η = 1 - 250/500 = 1 - 0.5 = 0.5. Step 4: Converting to percentage: η = 0.5 × 100 = 50%. Step 5: Option A (25%) is incorrect. Option B (75%) might come from wrong subtraction. Option D (100%) is impossible since T₂ ≠ 0 K. Only 50% is correct.

Two isothermal and two adiabatic processes

Step 1: The Carnot cycle consists of exactly four processes in sequence. Step 2: Process 1 (A→B): Isothermal expansion at temperature T₁ (heat Q₁ absorbed from source). Step 3: Process 2 (B→C): Adiabatic expansion (temperature drops from T₁ to T₂, no heat exchange). Step 4: Process 3 (C→D): Isothermal compression at temperature T₂ (heat Q₂ rejected to sink). Step 5: Process 4 (D→A): Adiabatic compression (temperature rises back from T₂ to T₁). So the cycle has 2 isothermal + 2 adiabatic processes. All other options are incorrect.

The sink temperature T₂ must be 0 K or source temperature T₁ must be infinite.

Step 1: The efficiency formula is η = 1 - T₂/T₁. Step 2: For η = 1 (100%), we need T₂/T₁ = 0. Step 3: This is possible only if T₂ = 0 K (absolute zero) OR T₁ → ∞. Step 4: Both conditions are practically impossible — we cannot achieve absolute zero temperature, and no source can be at infinite temperature. Step 5: Therefore, 100% efficiency is a practical impossibility. Option B is wrong (T₁ = 0 K would give η = 1 - T₂/0, which is undefined and means no work). Option C (T₁ = T₂) gives η = 0, not 100%.

15%

Step 1: Given: Work done W = 5.4 × 10⁸ J/min, Heat absorbed Q₁ = 3.6 × 10⁹ J/min. Step 2: Efficiency formula: η = W/Q₁. Step 3: η = (5.4 × 10⁸) / (3.6 × 10⁹) = 5.4/36 = 0.15. Step 4: η = 0.15 × 100 = 15%. Step 5: Option A (10%) is too low. Option C (20%) and D (25%) are too high. The calculation clearly gives 15%. This is close to the real efficiency of steam engines (10-17%).