Entropy Practice Problems [Thermal & Statistical Mechanics I Studied #25]

Prob 14.1

A mug of tea was left to cool from 90℃ to 18℃. If there is 0.2kg of tea in the cup and the specific heat capacity at constant pressure is 4200 J/K·kg, show that the entropy of the tea decreases by 185.7 J/K. Explain the sign of this result.

I’ll interpret this problem as the contact between a small system and a big reservoir.

Hand-drawn: a small circle labelled system inside a huge circle labelled Reservoir (insanely huge); a red arrow runs from the system to the note T₁ loses dQ₁ / reservoir still at T_R ···· dQ₁ in

The same drawing one step later: the system is now at T₂, and a second red arrow leads to T₂ loses dQ₂ / reservoir still at T_R ···· dQ₂ in

The same drawing with a third step, T₃ loses dQ₃ / reservoir still at T_R ···· dQ₃ in., and red dots for and so on

Since calculating the entropy of the system and calculating the entropy of the surroundings are slightly different, I tried drawing it as a picture.

So if I do the system entropy calculation first, the system has its temperature T changing from moment to moment: in $\int \frac{\delta Q}{T}$ , T keeeeps changing.

Now, the situation in the problem above is one where the pressure is constant, so the change in heat $\delta Q$ can be calculated as $(0.2\,\text{kg})\,C_p\,dT$. That is,

The tea’s entropy change: ΔS_system = integral from 90℃ to 18℃ of (0.2 kg) C_p dT / T = (0.2 kg)(4200 J/K·kg) times the integral from 363.15 to 291.15 of dT/T = 0.2 × 4200 ln(291.15/363.15) ≅ −185.505

If I also calculate the entropy of the reservoir, then when I go to do $\int \frac{\delta Q}{T}$ , the temperature T of the reservoir is constant at 18℃. So it can come out of the integral~

The reservoir side: integral of δQ/T = (1/T) times the integral of δQ; we can get the heat change the same way as for the system, and the heat the system loses is exactly the heat the reservoir gains; = (1/291.15)(0.2 × 4200 × (363.15 − 291.15)) ≅ 207.73

That is, the change in total entropy is $207.73 - 185.62 = 22.11$. Yesss!!! It doesn’t violate the second law of thermodynamics, $dS \geq 0$.

And why does the entropy of the system decrease, then?

The thing about entropy increasing,,…. that was when we assumed it was a thermally isolated system.

So for systems that aren’t thermally isolated, it doesn’t always have to be positive…!!!

OK OK

Prob 14.4

a) A bath of water initially at 20℃ is put in contact with a heat reservoir at 80℃. Find the change in entropy.

b) Find the change in entropy of the heat reservoir when process a) happens.

The bathroom and its contents have a heat capacity of $10^4$ J/K.

Should I read that as a bathtub? Anyway, the entropy change of the bathtub is

The bathtub’s entropy change: ΔS_system = integral of δQ/T = integral from 20℃ to 80℃ of C_p dT/T = C_p times the integral from 293.15 to 353.15 of dT/T = C_p ln(353.15/293.15) ≅ 1862.08

The bathroom (our reservoir) will be at 80℃ the whoooole time, so

The reservoir’s entropy change: ΔS_reservoir = integral of δQ/T = (1/T) times the integral of C_p dT = (10⁴/80℃)(−60) = −(1/353.15)·10⁴·60 = −1698.994 ≅ 1699.0

That is, the change in total entropy is $1862.08 - 1699.0 = 163$.

I did solve part c, but I’m not confident…. T_T lol

Prob 14.5

A lead block with a heat capacity of 1 kilo J/K is cooled from 200K to 100K in two ways.

a) Drop it into a large fluid at 100K.

b) Cool it down to 150K in one fluid at 150K, then cool it down to 100K in another fluid at 100K.

For both cases, calculate the entropy change of the system made up of the block and the baths, when it cools from 200K to 100K. Prove that in the limit of infinitely many intermediate baths, the change in total entropy is zero.

a) The entropy change of the lead block is

The lead block: integral from 200 to 100 of δQ/T = integral from 200 to 100 of C_p dT/T = C_p ln(1/2) = −10³ ln 2 ≅ −693.147

The entropy change of that fluid is

That is, the change in total entropy is $1000 - 693.147 = 306.853$ J/K.

Now I’ll do the cooling in two stages.

The entropy change of the lead in round 1

Round 1, the lead: ΔS_lead1 = integral from 200 to 150 of δQ/T = integral from 200 to 150 of C_p dT/T = 10³ ln(3/4) ≅ −287.682

The fluid entropy change in round 1

Round 1, the fluid at 150 K: ΔS_fluid1 = integral of δQ/150 = integral of C_p dT/150 = (10³/150)·ΔT = (10³/150)·50 ≅ 333.333

The entropy change of the lead in round 2

Round 2, the lead: ΔS_lead2 = integral from 150 to 100 of δQ/T = integral from 150 to 100 of C_p dT/T = 10³ ln(2/3) ≅ −405.465

The fluid entropy change in round 2

Round 2, the fluid at 100 K: ΔS_fluid2 = integral of δQ/100 = integral of C_p dT/100 = (10³/100)·ΔT = (10³/100)·50 = 500

Therefore the change in total entropy is

c) Now I’ll try doing it infinitely many times.

Looking at the entropy change of the lead, the integrals have this property of merging into one?!

Then if I set the temperature step between the fluids as $\varepsilon$: plop into $200-\varepsilon$, plop into $200-2\varepsilon$, plop into $200-3\varepsilon$, …., plop into $200-n\varepsilon$. Then the change in total entropy is

Handwritten sum ΔS_lead1 + ΔS_lead2 + ΔS_lead3 + ΔS_lead4 + … + ΔS_leadn + ΔS_fluid1 + … + ΔS_fluidn; in red under the lead terms: these guys can squish into the integral from 200 to 100 of dQ/T, and; in blue under the fluid terms: these guys? Forget merging, all split up

That is, the change in total entropy is

ΔS_total = integral from 200 to 100 of δQ/T + C_pε/(200−ε) + C_pε/(200−2ε) + C_pε/(200−3ε) + … + C_pε/(200−nε); nε eventually = 100, and then we’ll send n → ∞, ε → 0; = −C_p ln 2 + C_pε Σ 1/(200−kε)

ΔS_total = −C_p ln 2 + C_pε Σ 1/(200−kε); now n gets sent off to infinity: the limit as n → ∞ of ΔS_total = −C_p ln 2 + C_pε times the limit of Σ 1/(200−kε)

The Riemann sum from high-school calculus: let 200 − kε = x, then dx = −ε; k = 1 gives x = 200 (since ε → 0) and k = n gives x = 100 (since nε = 100); written as an integral the sum is C_p times the integral from 200 to 100 of (1/x)(−dx), so the limit of ΔS_total = −C_p ln 2 − C_p ln(1/2) = 0

Prob 14.7

$n$ moles of gas are confined at temperature $T$ and volume $V$. Derive the expression for the entropy change

a) when it becomes $\alpha V$ through a reversible isothermal expansion, and

b) when the partition is removed and it undergoes free expansion into a vacuum.

i) Isothermal expansion necessarily requires heat to be supplied.

Reversible isothermal expansion: TdS = pdV (since dU = 0); ΔS_system = integral of pdV/T = integral from V to αV of n_mR/V dV = n_mR ln α; it is reversible, so dQ = TdS and ΔQ = TΔS: the heat the surroundings gave is the system’s entropy gain times T, so ΔS_surrounding = −n_mR ln α and ΔS_universe = 0

ii)

Free expansion: the system’s entropy change is the same, n_mR ln α, because the volume still goes from V to αV; free expansion is already adiabatic, so δQ = 0 → dS = 0 and the surroundings’ entropy does not change, ΔS_sur = 0; so ΔS_universe = n_mR ln α

Heads-up. The figures here are the original notes, slips and all. A few things to watch for:

In the tea’s entropy change, $0.2 \times 4200 \ln\frac{291.15}{363.15} \approx -185.62$ (the picture’s $-185.505$ is off), so the total in the text is $207.73 - 185.62 \approx 22.11$.
In the bathtub reservoir’s entropy change, the temperature goes in as $353.15$ K (not 80 °C), and the result keeps its minus sign: $\approx -1699.0$.
In the limit pictures, $\varepsilon = 100/n$ shrinks with $n$, so it belongs inside the limit: $C_p \lim_{n\to\infty}\sum_k \frac{\varepsilon}{200 - k\varepsilon} = C_p\int_{200}^{100}\frac{1}{x}(-dx)$.

Prob 14.1

Prob 14.4

Prob 14.5

Prob 14.7

Comments