Question 16.6: Entropy and the Carnot cycle Let’s return to the Carnot engi...
Entropy and the Carnot cycle
Let’s return to the Carnot engine in Example 16.2 (Section 16.6). The engine takes 2000 J of heat from a reservoir at 500 K, does some work, and discards some heat to a reservoir at 350 K. Find the total entropy change in the engine during one cycle.
SET UP AND SOLVE During the isothermal expansion at 500 K, the engine takes in 2000 J and its entropy change is
\mathrm{\Delta S=\frac{Q}{T}=\frac{2000 J}{500 K} =4.0 J/K. }
During the isothermal compression at 350 K, the engine gives off 1400 J of heat and its entropy change is
\mathrm{\Delta S=\frac{-1400 J}{350 K}=-4.0 J/K. }
The entropy change of the engine during each of the two adiabatic processes is zero. Thus, the total entropy change in the engine during one cycle is 4.0 J/K + 0 – 4.0 J/K + 0 = 0.
REFLECT The total entropy change of the two heat reservoirs is also zero, although each individual reservoir has a nonzero entropy change.
This cycle contains no irreversible processes, and the total entropy change of the system and its surroundings is zero.
Practice Problem: If the Carnot engine absorbs 3000 J of heat from the hot reservoir, how much work does the engine do and what is the entropy change of the cold reservoir during the cycle? Answer: 900 J, 6.0 J/K.