Physics 4311: Thermal Physics – HW Solution 7

due date: Wednesday, Mar 18, 2026

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Problem 1: Heat pump (6 points)

a)

In class we found for Carnot engine running forward:

| \frac{W}{Q_{h}} | = \frac{- W}{Q_{h}} = \frac{Q_{h} + Q_{l}}{Q_{h}} = 1 - \frac{T_{l}}{T_{h}}

Efficiency of Carnot enegine running backwards as heat pump:

η_{HP} = \frac{- Q_{h}}{W} = \frac{T_{h}}{T_{h} - T_{l}}

Here, $T_{h} = 72^{\circ} F = 295.4$ K and $T_{l} = 40^{\circ} F = 277.6$ K

η_{HP} = 16.6

b)

$T_{h} = 72^{\circ} F = 295.4$ K and $T_{l} = 0^{\circ} F = 255.4$ K

η_{HP} = 7.4

Problem 2: Air conditioner (14 points)

a)

In steady state, the heat leaking into the house equals the heat removed by the air conditioner

Δ Q = A (T_{h} - T_{l}) = Q_{l}

To relate $Q_{l}$ to electrical work $E$ , start from first law $E + Q_{h} + Q_{l} = 0$ . For air conditioner, $Q_{l} > 0, Q_{h} < 0, E > 0$ . From class, we know $Q_{h} ∕ Q_{l} = - T_{h} ∕ T_{l}$ . Thus,

E + Q_{l} - Q_{l} \frac{T_{h}}{T_{l}} = E + Q_{l} \frac{T_{l} - T_{h}}{T_{l}} = 0

Q_{l} = \frac{T_{l}}{T_{h} - T_{l}} E

Inserting into energy balance

\frac{T_{l}}{T_{h} - T_{l}} E = A (T_{h} - T_{l})

Solve for $T_{l}$

T_{l} = T_{h} + \frac{E}{2 A} \pm \sqrt{{(T_{h} + \frac{E}{2 A})}^{2} - T_{h}^{2}}

Chose $-$ sign because $T_{l} < T_{h}$ .

b)

Start from energy balance

T_{l} \frac{E_{\max}}{2} = A {(T_{h} - T_{l})}^{2} T_{l} E_{\max} = A {(T_{h}^{'} - T_{l})}^{2}

\sqrt{2} (T_{h} - T_{l}) = (T_{h}^{'} - T_{l})

Here, $T_{h} - T_{l} = 1 8^{\circ}$ F. Thus, $T_{h}^{'} = 7 2^{\circ}$ F $+ 25 . 4^{\circ}$ F $= 97 . 4^{\circ}$ F.

Problem 3: Carnot process for a paramagnetic substance (20 points)

a)

isotherms are straight lines through the origin, $M = DH ∕ T$

Δ Q_{12} = - Δ W_{12} = - \int_{M_{1}}^{M_{2}} HdM = - \int_{M_{1}}^{M_{2}} (T_{1} M ∕ D) dM = - (M_{2}^{2} - M_{1}^{2}) T_{1} ∕ (2 D)

b)

to find adiabatic $H - M$ curves, start from $0 = δQ = CdT - HdM$
use equation of state to substitute for $H$ giving $CdT = HdM = (MT ∕ D) dM$
integrate ODE: $CDln (T_{2} ∕ T_{1}) = (M_{3}^{2} - M_{2}^{2}) ∕ 2$ or equivalently $T_{2} ∕ T_{1} = \exp [(M_{3}^{2} - M_{2}^{2}) ∕ (2 CD)]$
use equation of state to substitute for $T$ giving

H_{3} ∕ H_{2} = (M_{3} ∕ M_{2}) \exp [(M_{3}^{2} - M_{2}^{2}) ∕ (2 CD)]

c)

Sketch of Carnot cycle, consisting of two isothermal changes in $M$ and two adiabatic changes in $M$ in the $M - H$ plane.

Schematic of Carnot cycle in H-M phase diagram

d)

1–2:

isothermal at $T_{1}$ : $dU = 0$

Δ Q_{12} = - Δ W_{12} = - \int_{M_{1}}^{M_{2}} HdM = - \int_{M_{1}}^{M_{2}} (T_{1} M ∕ D) dM = - (M_{2}^{2} - M_{1}^{2}) T_{1} ∕ (2 D)

2–3:

adiabatic, $Δ Q_{23} = 0$

3–4:

isothermal at $T_{2}$ : $dU = 0$

Δ Q_{34} = - Δ W_{34} = - \int_{M_{3}}^{M_{4}} HdM = - \int_{M_{3}}^{M_{4}} (T_{2} M ∕ D) dM = - (M_{4}^{2} - M_{3}^{2}) T_{2} ∕ (2 D)

4–1:

adiabatic, $Δ Q_{41} = 0$

e)

from the equations of the two adiabatic curves: $(M_{2}^{2} - M_{3}^{2}) = (M_{1}^{2} - M_{4}^{2})$

η = 1 + \frac{Δ Q_{34}}{Δ Q_{12}} = 1 + \frac{(M_{3}^{2} - M_{4}^{2}) T_{2}}{(M_{1}^{2} - M_{2}^{2}) T_{1}} = 1 - \frac{T_{2}}{T_{1}}