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 U U, 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.
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