Chapter 13 Practice Problems [Thermal & Statistical Mechanics I Studied #21]

Ex 13.4: Efficiency of a refrigerator?!

A typical engine or system takes in heat from a hot reservoir, scarfs it down nicely, and the usual thing is to dump the rest into a (relatively) cold reservoir.

But a refrigerator is the opposite: it takes heat from a cold reservoir. (artificially….)

A hand-drawn refrigerator: an empty box wrapped in a frame of little electrical symbols (coil, capacitor, switch), with the title Refrigerator on top

Here’s a refrigerator, like so.

The same hand-drawn refrigerator with red arrows pulling the heat out of the inside

The refrigerator “does work.”

What kind of work does it do?

It does the work of pulling out the heat inside. By doing $W$ of work, it extracts $Q_\ell$ of heat and spits out an even bigger Q_h of heat, let’s say. If we sketch that out roughly

Sketch of the refrigerator as an engine: work W goes in, heat Q_ℓ comes in from the cold reservoir at T_ℓ and the bigger heat Q_h goes out to the hot reservoir at T_h

Alright, now let’s think about the efficiency. The efficiency here is what you got done over the work I did — this kind of ratio is what we should call efficiency, right?!

That is!!

Efficiency = what you got done (Q_ℓ of heat sucked out) over the work I did (W) = Q_ℓ/W = Q_ℓ/(Q_h − Q_ℓ); if this engine ran a Carnot cycle, Q_ℓ/(Q_h − Q_ℓ) = (Q_ℓ/Q_h)/(1 − Q_ℓ/Q_h) = (T_ℓ/T_h)/(1 − T_ℓ/T_h) = T_ℓ/(T_h − T_ℓ)

Prob 13.2

What is the maximum possible efficiency of an engine operating between two reservoirs?

One reservoir is at 100°C, the other reservoir is at 0°C. The strongest heat engine in the universe is the Carnot engine, and the thermal efficiency of the Carnot engine is

The Carnot efficiency: η_carnot = W/Q_h = (Q_h − Q_ℓ)/Q_h = (T_h − T_ℓ)/T_h = 1 − T_ℓ/T_h = 1 − 273.15/373.15 ≅ 0.267988 ≅ 27%

Prob 13.3

i) Perpetual motion machine of the first kind: it violates the first law of thermodynamics, the law of conservation of energy.

ii) Perpetual motion machine of the second kind: the Carnot engine, a heat engine that follows the Carnot cycle, has the strongest thermal efficiency in the universe.

Since even the strongest engine in the universe has less than 100% efficiency, a perpetual motion machine of the second kind, that is, a heat engine with 100% efficiency, does not exist, at least in the universe.

Prob 13.4

A possible ideal gas cycle operates as follows.

i) From the initial state $(p_1, V_1)$, the gas is cooled at constant pressure to $(p_1, V_2)$.

ii) The gas is heated at constant volume to $(p_2, V_2)$.

iii) The gas expands adiabatically and returns to $(p_1, V_1)$.

Assuming the heat capacity is constant (with respect to temperature), show that the efficiency is as follows.

The efficiency to show: 1 − γ((V_1/V_2) − 1)/((p_2/p_1) − 1)

Hand-drawn p-V diagram of the cycle: the corners (p_1, V_1) at T_1, (p_1, V_2) at T_2 and (p_2, V_2) at T_3 with dotted lines down to the V axis and across to the p axis, two black curves, and the red dotted adiabat from (p_2, V_2) to (p_1, V_1)

First, since this is a problem of finding efficiency, eta = heat used over heat in we need to head in the direction of getting this.

That means we need to calculate the heat taken in and the heat released!!!

Step i) is an isobaric contraction: (dQ/dT)_p = C_p, so dQ = C_p dT and ΔQ = C_p(T_2 − T_1), where the red bracket is negative (the red one is negative!), so ΔQ = −C_p(T_1 − T_2); that is, the heat change ΔQ here must mean the heat released

Step ii) is isochoric: (dQ/dT)_V = C_V, so dQ = C_V dT and ΔQ = C_V(T_3 − T_2); this much heat is the heat taken in

So the efficiency is η = (C_V(T_3 − T_2) − C_p(T_1 − T_2))/(C_V(T_3 − T_2)) = 1 − (C_p/C_V)(T_1 − T_2)/(T_3 − T_2) = 1 − γ(T_1 − T_2)/(T_3 − T_2)

$The fraction (T_1 − T_2)/(T_3 − T_2) needs the adiabatic condition of the third process, which we have not used yet: T_3 V_2^(γ−1) = T_1 V_1^(γ−1), so T_3 = T_1 (V_1/V_2)^(γ−1) (the fraction at the left of the picture has T_3 − T_2 in its numerator where T_1 − T_2 is meant)$

If we plug this in!!!!!

Plugging in: η = 1 − γ((T_1 − T_2)/(T_3 − T_2)) = 1 − γ((T_1 − T_2)/(T_1 (V_1/V_2)^(γ−1) − T_2)) = 1 − γ((1 − T_2/T_1)/((V_1/V_2)^(γ−1) − T_2/T_1)) = 1 − γ((1 − T_2/T_1)/((V_2/V_1)(V_1/V_2)^γ − T_2/T_1))

Here let me list out the ideal gas equation of state

The ideal gas equations at the three corners, p_1V_1 = n_mRT_1, p_1V_2 = n_mRT_2 and p_2V_2 = n_mRT_3, give T_2/T_1 = V_2/V_1; another form of the adiabatic process, p_2V_2^γ = p_1V_1^γ, gives (V_1/V_2)^γ = p_2/p_1, which I’ll plug in too

Substituting both relations: η = 1 − γ((1 − V_2/V_1)/((V_2/V_1)(p_2/p_1) − V_2/V_1)) = 1 − γ(((V_1/V_2) − 1)/((p_2/p_1) − 1)), QED

Prob 13.5

Show that the efficiency of the standard Otto cycle is $1-r^{1-\gamma}$ . (The Otto cycle is a 4-stroke cycle used in internal combustion engines for cars, trucks, etc.) Here $r$ is $r = \frac{V_1}{V_2}$ , the compression ratio.

Hand-drawn p-V diagram of the Otto cycle: the corners I, II, III and IV (the picture writes VI for the top right one), two adiabatic curves and two isochoric lines, with V_2 = V_3 and V_1 = V_4 marked on the V axis

This problem is also about calculating efficiency. Ⅰ → Ⅱ and Ⅲ → Ⅳ are adiabatic processes, so there’s no heat exchange, while in the isochoric process parts, Ⅱ → Ⅲ and Ⅳ → Ⅰ, there will be heat exchange.

the old equation editor doesn’t support Roman numerals so….. T_T I’ll go with Arabic numerals T_T

Step 2 → 3: since 2 → 3 is an isochoric process, let’s use the equation dQ = C_V dT; ΔQ = C_V(T_3 − T_2); T_3 is bigger than T_2, so the heat is positive: heat taken in

Step 2 → 3 again (the very same picture twice): since 2 → 3 is an isochoric process, let’s use the equation dQ = C_V dT; ΔQ = C_V(T_3 − T_2); T_3 is bigger than T_2, so the heat is positive: heat taken in

Step 4 → 1: also isochoric, so dQ = C_V dT and ΔQ = C_V(T_1 − T_4); T_1 is smaller than T_4, so the heat is negative: heat released; ΔQ = −C_V(T_4 − T_1)

So efficiency = heat used over heat in = (C_V(T_3 − T_2) − C_V(T_4 − T_1))/(C_V(T_3 − T_2)) = 1 − (T_4 − T_1)/(T_3 − T_2)

Now the adiabatic processes: process 1 gives T_1 V_1^(γ−1) = T_2 V_2^(γ−1), so T_1 = T_2 (V_2/V_1)^(γ−1); process 3 gives T_3 V_3^(γ−1) = T_4 V_4^(γ−1), so T_4 = T_3 (V_3/V_4)^(γ−1)

Alright, now let’s plug them in and unify things a bit.

Plugging in: efficiency = 1 − (T_4 − T_1)/(T_3 − T_2) = 1 − (T_3 (V_3/V_4)^(γ−1) − T_2 (V_2/V_1)^(γ−1))/(T_3 − T_2); since V_1 = V_4 and V_2 = V_3 this is 1 − (V_2/V_1)^(γ−1) = 1 − (V_1/V_2)^(1−γ) = 1 − r^(1−γ), QED (the picture leaves out the second equals sign)

Prob 13.6

An ideal air conditioner absorbs heat $Q_2$ from the house at temperature $T_2$ and releases heat $Q_1$ to the outside at temperature $T_1$, operating while consuming electrical energy $E$. The heat loss leaking from the house follows Newton’s law of cooling, $Q = A\left[T_1 - T_2\right]$ , where $A$ is a constant.

When steady state is reached, express $T_2$ in terms of $T_1$, $E$, $A$ for continuous operation… The air conditioner is controlled by a thermostat. It is designed so that, with the thermostat set to 20°C and the outside temperature at 30°C, it operates at 30% of its maximum electrical input.

What is the highest outside temperature at which the house can still be kept at 20°C?

This problem is kinda the same principle as the refrigerator we did earlier hehe

The same hand-drawn refrigerator with red arrows pulling the heat out of the inside

We can think of it as a refrigerator that uses energy $E$ to extract heat $Q_2$ at the house’s temperature $T_2$ and throws $Q_1$ to the outside at temperature $T_1$. Ah… since this is an air conditioner, I drew another picture.

Whisen, let’s go!

A hand-drawn wall-mounted air conditioner, Whisen, with red arrows pulling heat into the unit: eats heat

The same air conditioner with blue swirls of air coming out of the vent: a red arrow out at the top says dumps heat and the red arrows in at the bottom say eats heat

The air conditioner runs on energy $E$, that is, it uses energy $E$ to pull out $Q_2$ and throw heat to the outside, and since we’re told it’s an ideal air conditioner (engine), $\frac{Q_2}{Q_1} = \frac{T_2}{T_1}$ should hold!!!

Let’s start: E + Q_2 = Q_1, so E = Q_1 − Q_2 = Q_2 T_1/T_2 − Q_2 = Q_2 (T_1/T_2 − 1)

Here we need to consider the loss.

Diagram of the ideal air conditioner as an engine: the circle E takes in Q_2 from below (the house at T_2) and sends heat up to the outside (T_1)

In the middle of pulling the heat up

The same diagram with the red note: Yikes! Heat loss! A(T_1 − T_2), and an arrow bending away from the circle

Yikes!!! Heat loss occurs.

So in the end, we pull out $Q_2$ only to have that much leak right back in. That is, $Q_2 = A\left[T_1 - T_2\right]$ balances it out!!?

An equilibrium — well, a steady state — will be reached! Ah ah, the heat will balance out!!! That is!

E = Q_2 (T_1/T_2 − 1) = A(T_1 − T_2)(T_1/T_2 − 1) = (A/T_2)(T_1 − T_2)^2

If we first organize this equation,…

E T_2 = A(T_1 − T_2)^2 = A(T_1^2 − T_1T_2 + T_2^2), and then A T_2^2 − (2A T_1 + E) T_2 + T_1^2 = 0 (the picture has T_1T_2 where 2T_1T_2 is right in the first line, and T_1^2 where A T_1^2 is right in the second)

We can find $T_2$ with the quadratic formula, and express it in terms of $T_1$, $E$, $A$!!!

The quadratic formula: T_2 = ((2A T_1 + E) ± √((2A T_1 + E)^2 − 4A T_1^2))/(2A) (the discriminant should read (2A T_1 + E)^2 − 4A^2 T_1^2)

Since the air conditioner is designed at 0.3 of the maximum input for 20°C ($T_2$) and 30°C ($T_1$), we can get the relation between $E$ and $A$ from this!!!

E = (A/T_2)(T_1 − T_2)^2; at 0.3E: 0.3E = (A/293.15)(303.15 − 293.15)^2 = (A/293.15)(10)^2, so 0.87945E = A

For the maximum outside temperature at which the house can be kept at 20°C, we just need to find $T_1$ when the air conditioner runs at 100% of its maximum input.

At full input: E = (A/T_2)(T_1 − T_2)^2, so (1/0.87945)T_2 = (T_1 − T_2)^2 and T_1 − T_2 = √(T_2/0.87945); T_2 = 293.15 gogo; so T_1 = T_2 + √(T_2/0.87945) = 293.15 + √(293.15/0.87945) = 311.61 K ≅ 38.25 °C (the arithmetic gives 311.41 K, that is 38.26 °C)

Prob 13.7

Two identical objects with constant heat capacity $C_p$ serve as the two heat reservoirs of a heat engine, at temperatures $T_1$ and $T_2$. If the objects are under constant pressure, show that the obtainable work is as follows.

If the most efficient engine is used, T_f^2 = T_1T_2 holds.

Under constant pressure, the heat gained and the heat lost are

DQ = C_p dT; so dQ_h = C_p(T_f − T_h) = −C_p(T_h − T_f), heat lost by the reservoir and gained by the engine; dQ_ℓ = C_p(T_f − T_ℓ), heat gained by the reservoir and lost by the engine; the work is the difference between the heat gained and the heat lost: W = |dQ_h| − |dQ_ℓ| = C_p(T_h − T_f) − C_p(T_f − T_ℓ) = C_p(T_h + T_ℓ − 2T_f)

T_f^2 = T_hT_ℓ can be shown with Clausius’ inequality: in a Carnot engine the closed-loop integral of dQ/T is 0, so the integral of dQ_h/T_h plus the integral of dQ_ℓ/T_ℓ is 0, that is C_p ln(T_f/T_h) + C_p ln(T_f/T_ℓ) = 0, so ln(T_f/T_h · T_f/T_ℓ) = ln 1 and T_f^2 = T_hT_ℓ

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

In the picture that starts the massaging, the fraction to rewrite is $\frac{T_1 - T_2}{T_3 - T_2}$ from the efficiency line above; the picture writes $T_3 - T_2$ in the numerator too.
In the Otto-cycle diagram, the corner at the top right is IV, not VI.
In the air-conditioner algebra, $A(T_1 - T_2)^2 = A(T_1^2 - 2T_1T_2 + T_2^2)$, so the quadratic is $AT_2^2 - (2AT_1 + E)T_2 + AT_1^2 = 0$ and its discriminant is $(2AT_1 + E)^2 - 4A^2T_1^2$; the pictures drop the 2 in the first line and one $A$ in the constant term and in the discriminant.
In the last numerical picture, $T_1 = 293.15 + \sqrt{\frac{293.15}{0.87945}} = 311.41$ K, about 38.26 degrees Celsius; the picture writes 311.61 K.