Question

A Carnot engine whose low-temperature reservoir is at 19.1°C has an efficiency of 30.7%. By how much should the Celsius temperature of the high-temperature reservoir be increased to increase the efficiency to 52.0%?

Answer 1

Answer:

The temperature of the high-temperature source must increase in 12.23ºC.

Explanation:

For a Carnot engine, the efficiency is defined as:

$n = 1- T2/T1$

Where T2 and T1 are the low and the high-temperature sources respectively. Therefore for the value of T2 of 19.1ºC and the n equal to 30.7% (0.307), the T1 Temperature can be calculated as:

$n = T1/T1 - T2/T1$

$n = (T1-T2)/T1$

$T1.n = (T1-T2)$

$T1-T1.n =T2[\tex]

[tex] T1(1-n) =T2$

$T1 =T2/(1-n)$

$T1 = 19.1\ºC /(1-0.307)$

$T1 = 27.56\ºC$

Then for the new effciencie n' of 52% (0.52) the new temeperature T1' will be:

$T1' =T2/(1-n')$

$T1' = 19.1\ºC /(1-0.52$

$T1' = 39.79\º C$

Finally the increment of temperature is:

$AT1 =T1'-T1$

$AT1 =39.79\º C-27.56\º C$

$AT1 =12.23\º C$

$AT1 =12.23\º C$