Temperature — Question 7

Question 7

Equal temperature can coexist with different phase energy

At a fixed pressure, liquid and solid forms of an ideal substance coexist at $T_m=300\,\mathrm K$. For a $2.0\,\mathrm{kg}$ sample, the supplied phase-energy model at $T_m$ is $U=U_s+fm\ell$, where $f$ is liquid mass fraction and $\ell=100\,\mathrm{kJ/kg}$. Two equilibrium preparations have $f_A=.20$ and $f_B=.80$.

1. Find $U_B-U_A$. Does a temperature measurement distinguish the two liquid fractions?

2. Preparation A gains $40\,\mathrm{kJ}$ solely by melting while remaining at coexistence. Find its new fraction and temperature.

Question 7 - Solution

Equal temperature can coexist with different phase energy

(a) Method and calculation

$$U_B-U_A=(f_B-f_A)m\ell=.60(2)(100)=120\,\mathrm{kJ}.$$$$T_A=T_B=T_m=300\,\mathrm K.$$A single temperature does not fix the fraction of coexisting phases or the energy.

(b) Interpretation and further calculation

$$\Delta f=\frac{\Delta U}{m\ell}=\frac{40}{2(100)}=.20,\qquad f_f=.40.$$$$T_f=300\,\mathrm K.$$Because $0<f_f<1$, the specified coexistence model still applies; the energy changes phase fraction rather than temperature.

Physical check

This exercise supplies its phase-energy relation; quantitative calorimetry and latent heat are developed later in the chapter.