In a carnot engine, the temperature of reservoir is 527°C and that of sink is 200 K. If the work done by the engine when it transfers heat from reservoir to sink is 12000 kJ, the quantity of heat absorbed by the engine from reservoir is ______________ × 10⁶ J.

Solution

Physics sir · 4 steps

Turn 527°C into kelvin first, then the Carnot efficiency hands you the heat absorbed directly.

1. Kelvin first. Carnot efficiency only works with absolute temperatures. The source is at T₁ = 527 + 273 = 800 K and the sink is at T₂ = 200 K.
2. Efficiency. η = 1 - T₂/T₁ = 1 - 200/800 = 0.75. So the engine turns three quarters of the heat it absorbs into work.

   η = W/Q₁ ⇒ Q₁ = W/η

3. Put in numbers. W = 12000 kJ = 1.2 × 10⁷ J, so Q₁ = (1.2 × 10⁷)/0.75 = 1.6 × 10⁷ J.
4. So. Q₁ = 16 × 10⁶ J, so the blank is 16.

The common slip: 19. Using 527 directly instead of 800 K gives η ≈ 0.62 and Q₁ ≈ 19.3 × 10⁶ J, so about 19, not 16.

The idea to keep. For a Carnot engine, η = 1 - T₂/T₁ = W/Q₁. Always put both temperatures in kelvin before taking the ratio.

Answer from the NTA final answer key. Solution by PrepWiser.