Determine the efficiency of the Carnot cycle if the temperatures of the heater and refrigerator are 200 and 15 ° C

Determine the efficiency of the Carnot cycle if the temperatures of the heater and refrigerator are 200 and 15 ° C, respectively. How much should the heater temperature be raised to double the cycle efficiency?

Given:

t1 = 200 ° Celsius = 473 ° Kelvin – heater temperature in the Carnot cycle;

t2 = 15 ° Celsius = 288 ° Kelvin – the temperature of the refrigerator in the Carnot cycle.

It is required to determine n (%) – the efficiency of the cycle, as well as dt (degrees Celsius) – by how much it is necessary to increase the heater temperature in order for the efficiency of the cycle to double.

The efficiency of the Carnot cycle will be equal to:

n = (t1 – t2) / t1 = (473 – 288) / 473 = 185/473 = 0.39 or 39%.

According to the condition of the problem, we will increase the efficiency by 2 times: n1 = 2 * n = 78 or 0.78.

Then the heater temperature should be equal to:

t3 = t2 / (1 – n1) = 288 / (1 – 0.78) = 288 / 0.22 = 1309 ° Kelvin = 1036 ° Celsius.

Then:

dt = t3 – t1 = 1036 – 200 = 836 ° Celsius.

Answer: the efficiency of the cycle is 39%, in order to increase the efficiency by 2 times, it is necessary to increase the heater temperature by 836 °.