Question

With each cycle, a 2500.-W engine extracts \(2100 .\) J from a thermal reservoir at \(90.0^{\circ} \mathrm{C}\) and expels \(1500 .\) J into a thermal reservoir at \(20.0^{\circ} \mathrm{C}\). What is the work done for each cycle? What is the engine's efficiency? How much time does each cycle take?

Short answer

Answer: The work done for each cycle is 600 J, the engine's efficiency is approximately 28.6%, and each cycle takes 0.24 seconds.

Step-by-step solution

Calculate the work done per cycle

The energy conservation principle states that the work done per cycle is equal to the difference between the input heat energy and output heat energy: W = Q_in - Q_out W = 2100 J - 1500 J W = 600 J The work done for each cycle is 600 J.

Compute the engine's efficiency

The efficiency of the engine is defined as the ratio of the work done per cycle to the input heat energy: Efficiency = W / Q_in Efficiency = 600 J / 2100 J Efficiency = 2 / 7 The engine's efficiency is 2/7 or approximately 0.286 or 28.6%.

Calculate the time for each cycle

We know the power of the engine, which is 2500 W. We can calculate the time for each cycle using the formula: Time = Work done per cycle / Power Time = 600 J / 2500 W Time = 0.24 s Each cycle takes 0.24 seconds.