Work done by a Carnot engine operating between temperatures 127° C and 27° C is 2 kJ. The amount of heat transferred to the engine by the reservoir is :

Solution

Physics sir · 4 steps

Get the efficiency from the two temperatures in kelvin, then heat in is just work divided by efficiency.

1. Convert to kelvin. The source is at T₁ = 127 + 273 = 400 K and the sink is at T₂ = 27 + 273 = 300 K.

   η = 1 - T₂/T₁ = W/Q₁

2. Find the efficiency. η = 1 - 300/400 = 1/4. So only one quarter of the heat taken from the hot reservoir turns into work.
3. Heat taken in. Q₁ = W/η = (2 kJ)/(1/4) = 8 kJ. The other 6 kJ is rejected to the sink at 300 K.
4. So. The reservoir transfers 8 kJ of heat to the engine, which is option (a).

The idea to keep. For any Carnot engine, η = 1 - T₂/T₁ with temperatures in kelvin, and η = W/Q₁. Two equations, so knowing any one of W or Q gives the rest.

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