Question

A Carnot engine has an efficiency of 35.4%. It operates between constant-temperature reservoirs differing in temperature by 100 C°. In kelvins, what are (a)the temperature of the hot reservoir and (b) the temperature of the cold reservoir?

Answer 1

here,

Efficiency n of a Carnot engine is defined to be:

n = 1 - Tc / Th = (Th - Tc) / Th

where

Tc is the absolute temperature of the cold reservoir, and

Th is the absolute temperature of the hot reservoir.

In this case, given is n = 35.4 % and Th - Tc = 100 degree C

Notice that although temperature difference is given in °C it has same numerical value in Kelvins because magnitude of the degree Celsius is exactly equal to that of the Kelvin (the difference between two scales is only in their starting points).

Th = (Th - Tc) / n

Th = 100 / 0.354 = 282.5 K (rounded to closest number)

Tc = Th - 100 = 182.5 K

Lower temperature is Tc = 182.5 K

Higher temperature is Th = 282.5 K