During the Carnot cycle, the working substance receives 300 kJ of heat from the heater. Temperatures of the heater 400 K

During the Carnot cycle, the working substance receives 300 kJ of heat from the heater. Temperatures of the heater 400 K, refrigerator 200 K. Determine the work done by the working substance per cycle.

In general, this is not such an impossible task. To solve it, you just need a couple of formulas that you need to know.

So, at the very beginning of the solution, we find Q.

Q = 3 * 10 ^ 5 J.

Now let’s write out the values of T1 and T2.

T1 = 400 K.

T2 = 200 K.

We need to find A.

Now we must write out a general formula, from which we will find A.

η = A / Q = (1 – T2 / T1).

Well, now we find directly the value of A.

A = Q * (1 – T2 / T1) = 3 * 10 ^ 5 (1 – 200/400) = 1.5 * 10 ^ 5 J.

Thus, we have solved this problem, having received the correct answer.