Show that for a closed loop in a Carnot engine, the energy is conserved in the heat flow at a certain temperature in each reservoir, i.e. show that #q_H/T_H + q_C/T_C = 0#?

A Carnot engine can be represented as follows:

Let me expand what is on the right.

#ointdS = int_(a)^(a) dS = S_i - S_i = 0#,

since #S# is entropy, which is a state function describing the amount of energy dispersal, and a closed integral, "#oint#", indicates that the initial and final states are identical (hence, #DeltaS = S_i - S_i = 0#).

By definition,

#dS = (deltaq_(rev))/T#,

where #q_(rev)# is reversible heat flow, i.e. heat flow applied in an infinitesimally slow manner, such that the system has the chance to re-equilibrate the entire time the heat is applied.

The hot reservoir, with temperature #T_H#, transfers heat #q_H > 0# (with respect to the engine) into the engine, and the engine transfers heat #q_C < 0# (with respect to the engine) out into the cold reservoir with temperature #T_C#.

If this heat transfer is exact, i.e. if we have conservation of energy and the heat is transferred reversibly, then:

#oint(deltaq_(rev))/T = ointdS = int (deltaq_H)/(T_H) + int (deltaq_C)/(T_C) = 0#

Hence, integrating with respect to the heat flow in from/to each reservoir, we have:

#color(blue)(q_H/T_H + q_C/T_C = 0)#