20.9.2 Entropy of Heating

Equation 20.51, $d S = d Q / T$ , is for an infinitesimal heating at temperature T. We can find the entropy change for a finite heat transfer, for which T changes, by integrating.

A photo of a boiling kettle releasing vapor pressure through the nozzle. — The volume of a gas is much larger than the volume of its liquid, so boiling increases the number of microstates—the number of ways that the molecules can be arranged to share the energy. A liquid-to-gas phase change is always accompanied by a large increase in entropy.

You learned in Chapter 19 that the heat required to change the temperature of a solid or liquid of mass M with specific heat c is $Q = M c Δ T$ . For an infinitesimal heating, $d Q = M c d T$ and thus $d S = M c d T / T$ . We can integrate from the initial entropy $S_{i}$ at initial temperature $T_{i}$ to the final values $S_{f}$ at $T_{f}$ . If we assume that the specific heat remains constant—usually a good assumption if the temperature change is modest—then

Δ S_{heat} = \int_{S_{i}}^{S_{f}} d S = M c \int_{T_{i}}^{T_{f}} \frac{d T}{T} = M c (ln T_{f} - ln T_{i}) = M c ln (\frac{T_{f}}{T_{i}})

(20.54)

EXAMPLE 20.11 Heating water

In Example 20.9 we found that the entropy of 100 g of water increases by $122 J / K$ in changing from 0°C ice to 0°C liquid water. To what temperature does the water need to be heated to increase its entropy by an additional $122 J / K ?$

MODEL Assume that the heating is carried out slowly.

SOLVE We can solve Equation 20.54 for $T_{f}$ by first dividing through by $M c$ , then using $e^{ln a} = a :$

T_{f} = T_{i} e^{Δ S / M c} = (273 K) e^{(122 J / K) / (0.100 kg) (4190 J / kg K)} = 365 K = 92 ° C

REVIEW A substantial temperature increase is needed to match the entropy increase of melting. The water’s entropy increases with temperature because faster molecules have more ways to disperse the kinetic energy.

In Chapter 19 an adiabatic process was defined as one in which no heat is transferred: $Q = 0$ . We can see from Equation 20.51, $d S = d Q / T$ , that there is no entropy change in an adiabatic process; $Δ S_{adiabatic} = 0$ . An adiabatic process is an isentropic process, one with constant entropy.