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 Q_0 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 system is found to be _ net = W Q_0 = 211 243 . The value of is:
A. A
Let Q_0 be the heat absorbed by the first engine and Q_1 be the heat rejected. The efficiency of the first engine is = 1 - Q_1 Q_0 Q_1 = Q_0(1 - ) Similarly, for the second engine, the heat rejected is Q_2 = Q_1(1 - ) = Q_0(1 - )^2 Following this pattern, for the 5^ th engine, the heat rejected is Q_5 = Q_0(1 - )^5 The total work done by all five engines is W = W_1 + W_2 + W_3 + W_4 + W_5 = (Q_0 - Q_1) + (Q_1 - Q_2) + + (Q_4 - Q_5) = Q_0 - Q_5 The net efficiency of the system is _ net = W Q_0 = Q_0 - Q_5 Q_0 = 1 - Q_5 Q_0 Substituting the value of Q_5, we get _ net = 1 - (1 - )^5 Given _ net = 211 243 1 - (1 - )^5 = 211 243 (1 - )^5 = 1 - 211 243 = 32 243 (1 - )^5 = ( 2 3 )^5 1 - = 2 3 = 1 3 Answer: 0.33
Related: Physics — Thermodynamics · All PYQ Banks