Advanced Thermodynamics   Quiz #4

(Open handouts & note sheet)

  1. We've seen videos of lava boiling the oceans off Hawaii. What weight of liquid iron, Fe, at its melting point (1538°C) will bring a liter of 25°C water to its boiling point?

    Species Average CP
    J/mol K     		Enthalpy of Fusion
    kJ/mol     		Enthalpy of Vaporization
    kJ/mol
    Fe(s)37.613.81-
    H2O(l)75.46.0150.65

    CP is assumed averaged over the relevant temperature range. A liter of water weighs 1 kg.

Since we can neither create nor destroy energy, the energy gained by the water in warming from 25°C to 100°C must have been lost by the iron in condensing at 1538°C then cooling to 100°C. So we need only set up that equivalence and solve for the only unknown, moles of Fe.

nH2O CP, H2O Delta TH2O = nFe [ Delta Hfus, Fe + CP, Fe Delta TFe]

nFe = nH2O CP, H2O Delta TH2O / [ Delta Hfus, Fe + CP, Fe Delta TFe]

nFe = 1000 gH2O×[1 molH2O / 18 gH2O]× 75.4 J/K molH2O×(100-25) K
        ÷ [ 13,810 J/molFe + 37.6 J/K molFe×(1538-100) K]

nFe = 4.63 molFe

Weight Fe = 4.63 molFe × 55.845 gFe/molFe = 258 gFe

which ought to surprise you. As hot as liquid iron is, the quenching power of water with its high heat capacity (twice that of iron on a molar basis... and over 6× higher on a weight basis!) is impressive. But then blacksmiths have known this since the Bronze Age.

    1. By what factor can an ideal diatomic gas expand reversibly and adiabatically starting at 25°C before it reaches the freezing point of water? (Assume that neither vibrations nor electronic states play any role in the heat capacities. A good assumption.)

      This can tell you something about condensation of water in rising air currents in our summer thunderstorms. Air, after all, is an ideal diatomic gas; two even! And its expansion in updrafts is nearly adiabatic. In fact, the change in T with altitude is called the "adiabatic lapse rate."

    • The expression we need to relate volume and temperature in a reversible, adiabatic expansion (or compression for that matter) is given below in part (b), viz.,

    T VR/CV = c

    T1 V1R/CV = T2 V2R/CV

    (V2/V1)R/CV = T1/T2

    V2/V1 = (T1/T2)CV/R

    Now for a diatomic molecule with no vibration or electronic states of thermal consequence, CV = CVTRANS+CVROT = 3(½R)+2(½)R = (5/2)R, so CV/R = 2.5.

    V2/V1 = (T1/T2)2.5 = (298/273)2.5 = 1.25

  • I warned you in class that the constants we tossed about in the expressions for reversible, adiabatic expansions weren't the same. Show that by determining c" in terms of c, where

    T VR/CV = c

    P VCP/CV = c"

    (Remember, we're talking about a mole of ideal gas here, but it doesn't have to be diatomic.)

    • Since it's a mole of ideal gas, we can eliminate T from the first equation using T=PV/R to give

    (PV/R) VR/CV = [P V(1 + R/CV)] / R = [P V(CV+R)/CV] / R = [P VCP/CV] / R = c

    P VCP/CV = c R = c"
    Last modified 22 October 1997.