Spontaneous heat transfer from hot to cold is an irreversible process. Calculate the total change in entropy if 4000 J of heat transfer occurs from a hot reservoir at T_h = 600 K(327°C) to a cold reservoir at T_c = 250 K(-23°C), assuming there is no temperature change in either reservoir. (See Figure 15.33.)

Strategy

How can we calculate the change in entropy for an irreversible process when ΔS_tot = ΔS_h + ΔS_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.

Data given in Figure 15.33

Transcribed Image Text:

T Q Direct from Tn to Te T Irreversible AS irrev = ASrev (a) Th Q Reversible process Reversible process = Q T₂ Two reversible processes AS irrev AS rev (b)