Since the change in entropy is Q / T Q / T, there is a larger change at lower temperatures. It is reasonable that entropy increases for heat transfer from hot to cold. (b) The same final state and, thus, the same change in entropy is achieved for the objects if reversible heat transfer processes occur between the two objects whose temperatures are the same as the temperatures of the corresponding objects in the irreversible process. First, for the heat transfer from the hot reservoir,įigure 15.33 (a) Heat transfer from a hot object to a cold one is an irreversible process that produces an overall increase in entropy. We now calculate the two changes in entropy using Δ S tot = Δ S h + Δ S c Δ S tot = Δ S h + Δ S c. This produces the same changes in the hot and cold reservoirs that would occur if the heat transfer were allowed to occur irreversibly between them, and so it also produces the same changes in entropy. So we can calculate the change in entropy of the hot reservoir for a hypothetical reversible process in which 4000 J of heat transfer occurs from it then we do the same for a hypothetical reversible process in which 4000 J of heat transfer occurs to the cold reservoir. How can we calculate the change in entropy for an irreversible process when Δ S tot = Δ S h + Δ S c Δ S tot = Δ S h + Δ S c is valid only for reversible processes? Remember that the total change in entropy of the hot and cold reservoirs will be the same whether a reversible or irreversible process is involved in heat transfer from hot to cold. Calculate the total change in entropy if 4000 J of heat transfer occurs from a hot reservoir at T h = 600 K 327º C T h = 600 K 327º C to a cold reservoir at T c = 250 K − 23º C T c = 250 K − 23º C, assuming there is no temperature change in either reservoir. Spontaneous heat transfer from hot to cold is an irreversible process.

(See Figure 15.32.)Įntropy Increases in an Irreversible (Real) Process That will be the change in entropy for any process going from state 1 to state 2.

We just need to find or imagine a reversible process that takes us from state 1 to state 2 and calculate Δ S Δ S for that process. Thus the change in entropy Δ S Δ S of a system between state 1 and state 2 is the same no matter how the change occurs. The reason is that the entropy s s of a system, like internal energy E int E int, depends only on the state of the system and not how it reached that condition. However, we can find Δ S Δ S precisely even for real, irreversible processes. The definition of Δ S Δ S is strictly valid only for reversible processes, such as used in a Carnot engine. If temperature changes during the process, then it is usually a good approximation (for small changes in temperature) to take T T to be the average temperature, avoiding the need to use integral calculus to find Δ S Δ S.

The SI unit for entropy is joules per kelvin (J/K). Where Q Q is the heat transfer, which is positive for heat transfer into and negative for heat transfer out of, and T T is the absolute temperature at which the reversible process takes place.

Increase in entropy