Entropy Increases in an Irreversible (Real) Process

How can we calculate the change in entropy for an irreversible process when $\mathrm{\Delta }{S}_{\text{tot}}=\mathrm{\Delta }{S}_{\text{h}}+\mathrm{\Delta }{S}_{\text{c}}$ 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. 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. 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.

We now calculate the two changes in entropy using $\mathrm{\Delta }{S}_{\text{tot}}=\mathrm{\Delta }{S}_{\text{h}}+\mathrm{\Delta }{S}_{\text{c}}$. First, for the heat transfer from the hot reservoir,

And for the cold reservoir,

Thus the total is

There is an increase in entropy for the system of two heat reservoirs undergoing this irreversible heat transfer. We will see that this means there is a loss of ability to do work with this transferred energy. Entropy has increased, and energy has become unavailable to do work.

Figure 15.34 (a) Heat transfer from a hot object to a cold one is an irreversible process that produces an overall increase in entropy. (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.