Reversible cycle entropy changes

Because the gas in the Carnot cycle starts and ends at the same state, the system s entropy does not change during a cycle. Now apply the second law to the universe for the case of the Carnot cycle. Because the processes are reversible, the entropy of the universe does not change by Equation 2b. This can be written ... 

In a Carnot s cycle, the entropy Qi/Ti is taken from the hot reservoir, and the entropy Q2/T2 is given up to the cold reservoir, and no other entropy change occurs anywhere else. Since these two quantities of entropy are equal and opposite, the entropy. change in the hot reservoir is exactly balanced, or, to use an expression of Clausius, is compensated by an equivalent change in the cold reservoir. Again, in any reversible cycle there is on the whole no production of entropy so that all the changes are compensated. 

A corollary to this law is that all reversible Carnot cycles, operating between the same two temperatures, must have the same efficiency. For a perfectly reversible cycle in which the only pressure, P, is uniform and external, then dU = TdS - PdU where df/ is the differential change in internal energy, T is the absolute temperature, dS is the differential change in entropy, and dV is the differential change in volume. This relation, which is a... 

Let S be the entropy and be the reversible heat added. In a given differential step, the heat added is dQ The differential change in entropy in every differential step is therefore dS = dQ JI. Around the cycle, the change in entropy is the integral, thus... 

By analyzing the Carnot cycle description of macroscopic energy transfer processes, Clausius demonstrated that the quantity J(l/T)dqrev is a state function, because its value for any reversible process is independent of the path. Based on this result, Clausius defined the procedure for calculating the entropy change AS = Sf — S for a system between any thermodynamic states i and f as... 

Fig. 24.2 Transfer of entropy 5t in a reversible cycle from a cold to a warm reservoir. Changes of volume are indicated by arrows (initial state contour line solid, final state contour line dashed). More heat Q flows off with the entropy St than in Qout > l2m. even though the body completely reverts to its initial state after every cycle and does not cool down at all. This means that energy is emitted as heat, which was not present in that form before but is generated. The question remains what phase of the process does this happen in and how ...

In the present proposition we seek to prove that /S is a function of state. Consider first of all the type of reversible cycle already discussed, consisting of two isot) ermals Ti and Tg o adiabatics. The entropy changes along the isothermals are obtained from (1 13) and are... 

Camot derived these principles from the abstract reversible cycle now called the Camot cycle. He assumed the validity of the caloric theory (heat as an indestmctible substance), which requires that the net heat in the cycle be zero, whereas today we would say that it is the net entropy change that is zero. 

Asa consequence of the additivity of (substitute reversible (19)) changes of heat entropy, when the medium C has gone through the reversible cycle O once, we have the following for the change AS g of the heat entropy Sc in the system AB) (consisting of the heater A and the cooler B) ... 

If the temperature remains fixed, it follows from (3.3.3) that for a reversible flow of heat Q, the change in entropy is QjT. In terms of entropy, Carnot s theorem (3.2.3) becomes the statement that the sum of the entropy changes in a reversible cycle is zero ... 

On the first isothermal branch a quantity of heat, Qi, Hows reversibly from the heat reservoir to the engine, both being a Ti similarly, on the second isothermal branch a quantity of heat Q2 Hows from the engine to the heat reservoir, both at T2. A total amount of work W is done in one cycle. For the isothermal steps we have the entropy changes... 

Another unbalanced energy component is quantitatively related to entropy changes of water associated with either sub-pressure development or relaxation within each sarcomere throughout an isometric cycle of hydraulic compression. This bulk baro-entropic reversible heat production cancels out during closed cycles, and it accounts to the thermo-elastic heat [87]- The baro-entropic component is quantitatively related to the coefficient of thermal water expansion. It is therefore predicted to change sign above and below 4°C, as observed. 

Example 3.4 uses the fact that the total entropy change of a reversible cycle is equal to zero to calculate the work produced by it. 

Before we attempt to calculate the entropy changes for the system and its surroundings, we must think about how to carry out the desired change in a reversible way. Because the water is below its normal freezing point, we cannot carry out this process in a reversible way by keeping the temperature at 263.15 K. To carry out the process in a reversible way, we must first carefully raise the temperature of the water to 273.15 K. At this temperature, water freezes reversibly. Once all the water has frozen, we slowly lower the temperature back to 263.15 K. The following thermodynamic cycle summarizes how super-cooled water can be converted to ice at 263.15 K in a reversible way. The overall process is shown in black and the reversible path representing the overall process is shown in blue. 

We wish to calculate the entropy change for the Carnot cycle, the reversible process by which we converted heat into work as illustrated in Figure 2.17. We then analyze an irreversible cyclic process that has definite directionality. Since this is a cyclic process in which the system returns to its initial state, all properties must return to their initial value, that is, for the system Aucyck = 0, and similarly As cycle = 0. The entropy change to the surroundings for the entire cycle equals the entropy change of each of the four steps ... 

For both Equations (3.7) and (3.4) to be true, the total entropy change to the surroundings for the four reversible processes in the Carnot cycle must be zero ... 

---