As shown in the figure, five Carnot engines, each with efficiency and same number of cycles per unit time, are operating between six heat reservoirs. The amount of heat released per cycle by one engine is completely absorbed by the next engine. Consider Q₀ to be the amount of heat absorbed per cycle by the first engine and W as the amount of total work done by all the engines per cycle, then the net efficiency of the

Step-by-step solution

Let Q₀ be the heat absorbed by the first engine and Q₁ be the heat rejected. The efficiency of the first engine is = 1 - Q₁ Q₀ Q₁ = Q₀(1 - ) Similarly, for the second engine, the heat rejected is Q₂ = Q₁(1 - ) = Q₀(1 - )^2 Following this pattern, for the 5^ th engine, the heat rejected is Q₅ = Q₀(1 - )^5 The total work done by all five engines is W = W₁ + W₂ + W₃ + W₄ + W₅ = (Q₀ - Q₁) + (Q₁ - Q₂) + + (Q₄ - Q₅) = Q₀ - Q₅ The net efficiency of the system is _ net = W Q₀ = Q₀ - Q₅ Q₀ = 1 - Q₅ Q₀ Substituting the value