Question

Consider a Carnot’s cycle operating between T₁ = 500 K and T₂ = 300 K producing 1 k J of mechanical work per cycle. Find the heat transferred to the engine by the reservoirs.

Answer 1

Carnot theorem: The efficiency of Carnot’s heat engine depends only on the temperature of source (T₁) and temperature of the sink (T₂), and heat supplied (Q₁) i.e., η = W/ Q₁ = 1 – T₂/T₁ (The efficiency of an engine is defined as the ratio of work done to the heat supplied.)

Carnot stated that no heat engine working between two given temperatures of source and sink can be more efficient than a perfectly reversible engine (Carnot engine) working between the same two temperatures. Carnot’s reversible engine working between two given temperatures is considered to be the most efficient engine.

According to the problem, the temperature of the source T₁ = 500 K,

The temperature of the sink T₂ = 300 K,

Work done per cycle W = 1 kJ = 1000 J

Heat transferred to the engine per cycle Q₁ = ?

The efficiency of a Carnot engine

`(η) = 1 - T_2/T_1`

= `1 - 300/500`

= `200/500`

= `2/5`

And `η = W/Q_1`

⇒ `Q_1 = W/η`

= `1000/(2/5)`

= 2500 J

As `Q_1 - Q_2 = W, Q_2 - Q_1 = W`

= 2500 J – 1000 J

= 1500 J