General

Give two statements of the second law of thermodynamics that are not obviously identical in conceptual content but are in fact equivalent.

Solution:

- Concepts:

The second law of thermodynamics - Reasoning:

We are supposed to express the second law of thermodynamics in two not obviously identical ways. - Details of the calculation:

(1) Heat cannot be taken in at a certain temperature with no other change in the system and converted into work.

(2) The total entropy of a closed system is always increasing.

Determine the work that can be obtained from the one cycle of the ideal
Carnot machine in which the working substance is a photon gas. The energy
density of this gas is given by the u = σT^{4 }and the pressure is P =
⅓u.

Solution:

- Concepts:

The second law of thermodynamics - Reasoning:

For an ideal, reversible engine Q_{1}/T_{1 }= Q_{2}/T_{2}. - Details of the calculation:

For a Carnot engine we have

Q_{1}/T_{1 }= Q_{2}/T_{2},

W_{max }= (Q_{high }- Q_{low})_{reversible }= Q_{high }- Q_{high}T_{low}/T_{high }= Q_{high}(1 - T_{low}/T_{high}).

This is the same for all Carnot engines and does not depend on the working fluid.

What is the efficiency of a Carnot cycle operating between the same high (T_{H})
and low (T_{C}) temperatures as the ideal gas in this circular cycle?

Solution:

- Concepts:

The Carnot cycle, the ideal gas law - Reasoning:

The efficiency of a Carnot engine is (T_{high}– T_{low})/T_{high}.

We use the diagram (geometry) and the ideal gas law to find T_{H}and T_{C}. - Details of the calculation:

For the circular cycle we have (P/P_{0}– 2)^{2}+ (V/V_{0}– 2)^{2}= 1. (equation of a circle)

For T_{H}and T_{C}we have P/P_{0}= V/V_{0}. (straight line)

Therefore at those temperatures 2(P/P_{0}– 2)^{2}= 1, P/P_{0}= 2 + 2^{-½}, for T_{H}and P/P_{0}= 2 - 2^{-½}, for T_{L}.

Using the ideal gas law, PV = nRT we have T_{H}= [P_{0}V_{0}/(nR)][ 2 + 2^{-½}]^{2}and T_{C}= [P_{0}V_{0}/(nR)][ 2 - 2^{-½}]^{2}.

The efficiency of a Carnot cycle operating between T_{H}and T_{C}is

eff = (T_{H}– T_{C})/T_{H}= [(2 + 2^{-½})^{2}- ( 2 - 2^{-½})^{2}]/(2 + 2^{-½})^{2}= 77.19%

__Heat engines__

An ideal heat engine is powered by two reservoirs of equal heat capacity C,
which is temperature independent. As the engine works, the reservoirs gradually
equilibrate.

(a) Find the overall efficiency of the engine from the starting point where the
reservoirs are at temperatures T_{1} = 90^{ o}C and T_{2}
= 30^{ o}C to the moment of complete equilibration.

(b) Now assume that a real heat engine is powered by the same reservoir with
initial temperatures T_{1} = 90^{ o}C and T_{2} = 30^{
o}C. Let C = 10^{5 }J/^{o}C. The real engine's overall
efficiency is 20% of the overall efficiency of the ideal engine. Find the
change in entropy of the system once the reservoirs have equilibrated.

A stretched rubber band contracts when heated under constant tension. Its
temperature increases when stretched adiabatically. The equation of state for
an idealized rubber band is J = αLT, where J is the tension in the rubber band,
L is its length, T is the absolute temperature and α is a constant.

For reversible processes we have for the rubber band TdS = dQ_{rev} = c_{L}dT
– JdL.

The heat capacity of the band at constant length is c_{L} = constant.

Consider a heat engine that uses a rubber band in the three-step cycle
shown.

Start with a stretched rubber band of length L_{0} , tension J_{0},
and temperature T_{A}.

Take the band through a sequence of reversible processes.

A --> B:

The rubber band is stretched under constant tension J_{0} to a length 2L_{0}
while in contact with a heat reservoir of temperature T_{B}.

B --> C:

While in contact with a heat reservoir of temperature T_{C} the tension
of the rubber band is increased from J_{0} to 2J_{0} at constant
length 2L_{0} .

C --> A:

While in contact with a heat reservoir of temperature T_{A}, tension and
length decrease linearly from (2J_{0}, 2L_{0}) to (J_{0},
L_{0}).

(a) Find the ratios T_{B}/T_{A} and T_{C}/T_{A}.
What is the ratio T_{hot}/T_{cold}?

(b) Find the work done by the heat engine as it moves through one cycle A --> B
--> C --> A.

(c) During one cycle A --> B --> C --> A, how much heat is extracted from the
hot reservoir and how much heat is dumped into the cold reservoir?

(d) Find the efficiency of this rubber-band heat engine and compare it to the
efficiency of a Carnot engine operating between the same temperatures.

Solution:

- Concepts:

A simple heat engine, reversible processes - Reasoning:

We are asked to take a simple heat engine through one cycle. - Details of the calculation:

(a) A --> B:

J_{A}= J_{0}= αL_{0}T_{A}, J_{B}= J_{0}= α2L_{0}T_{B}, T_{B}= T_{A}/2, ΔT = T_{B}– T_{A}= -T_{A}/2.

B --> C:

J_{B}= J_{0}= α2L_{0}T_{B}, J_{C}= 2J_{0}= α2L_{0}T_{C}, T_{C}= 2T_{B}= T_{A}, ΔT = T_{C}– T_{B}= T_{A}/2.

C --> A:

ΔT = 0.

T_{hot}= T_{A}, T_{cold}= T_{B}, T_{cold}/T_{hot}= ½.

(b) A --> B:

ΔW = -J_{0}L_{0}= -αL_{0}^{2}T_{A}.

B --> C:

ΔW = 0.

C --> A:

ΔW = -∫_{2L0}^{L0}JdL = -αT_{A}∫_{2L0}^{L0}LdL = (3/2)αL_{0}^{2}T_{A}.

ΔW_{tot}= -αL_{0}^{2}T_{A}+ (3/2)αL_{0}^{2}T_{A}= ½αL_{0}^{2}T_{A}.

(c) A --> B:

ΔQ = -c_{L}T_{A}/2 - αL_{0}^{2}T_{A}.

B --> C:

ΔQ = c_{L}T_{A}/2.

C --> A:

ΔQ = +(3/2)αL_{0}^{2}T_{A}.

ΔQ_{added }= (3/2)αL_{0}^{2}T_{A}+ c_{L}T_{A}/2.

ΔQ_{dumped }= (c_{L}T_{A}/2 + αL_{0}^{2}T_{A}).

(d) Efficiency = ΔW_{tot}/ΔQ_{added }= 1 /(c_{L}/(αL_{0}^{2}) + 3)

The efficiency of a Carnot engine operating between T_{hot}= T_{A}and T_{cold}= T_{B}is 1 - T_{B}/T_{A}= ½.

eff(Carnot)/eff(rubber band) = c_{L}/(2 αL_{0}^{2}) + 3/2.

A Carnot heat engine has the following entropy-temperature diagram.

(a) Describe the cycle. For each segment identify the
process, say whether work is done by the working system or on it and whether
heat is added to the system or extracted from it.

(b) How much work is done by the system?

Solution:

- Concepts:

The Carnot cycle, isothermal and adiabatic processes, entropy - Reasoning:

The Carnot cycle consists of isothermal expansion, adiabatic expansion, isothermal compression, and adiabatic compression. - Details of the calculation:

(a) A --> B: Isothermal expansion: The engine is in contact with a heat reservoir at temperature T_{2}. The engine does work. ΔW = ∫PΔdV. Heat is added to the working fluid.

B --> C: Adiabatic expansion: During the expansion the temperature falls to T_{1}.

The engine does work. ΔW = ∫PdV. No heat is added or extracted from the working fluid.

C --> D: Isothermal compression: The engine is in contact with a heat reservoir at temperature T_{1}. Work is done on the engine. ΔW = -∫PdV. Heat is extracted from the working fluid.

D --> A: Adiabatic compression: During the compression the temperature rises to T_{1}. Work is done on the engine. ΔW = -∫PdV. No heat is added or extracted from the working fluid.

(b) W = ΔQ_{high}- ΔQ_{low}= ΔS*T_{2}– ΔS*T_{1}= (S_{2}– S_{1})*(T_{2}– T_{1}).

The operation of a gasoline engine is (roughly) similar to the Otto cycle. A S-V diagram is shown.

A → B: Gas compressed adiabatically.

B → C: Gas heated
isochorically (constant volume; corresponds to combustion of gasoline).

C →
D: Gas expanded adiabatically (power stroke).

D → A: Gas cooled
isochorically.

Compute the efficiency of the Otto cycle for an ideal gas
(with temperature-independent heat capacities) as a function of the compression
ratio V_{A}/V_{B}, and the heat capacity of the gas, C_{V}
.

Solution:

- Concepts:

Heat engines - Reasoning:

Efficiency = net work done by the engine divided by the heat absorbed. We use the ideal gas law and the definition of the heat capacity C_{V}to find these quantities. - Details of the calculation:

Energy conservation: dU = dQ – PdV, dS = dQ/T, dU = TdS – PdV.

For an ideal gas we have PV = Nk_{B}T, C_{V}dT = dU.

Work done by the engine during one cycle: W_{tot}= W_{A-->B}+ W_{C-->D}= ∫_{A-->B}PdV + ∫_{C-->D}PdV.

Heat absorbed by the engine during one cycle: Q_{in }= Q_{B-->C }= ∫_{B-->C}TdS.

Efficiency: eff = W_{tot}/Q_{in}._{ }Finding W_{tot}:

W_{A-->B}= ∫_{A-->B}PdV = -Nk_{B}∫_{VB}^{VA}(T/V)dV.

S = S_{2}= constant along the path. We need to express T in terms of S and V.

dS = dU/T + PdV/T = C_{V}dT/T + Nk_{B}dV/V.

This integrates to S = C_{V}lnT + Nk_{B }lnV + constant.

Solving for T we find: T = α*exp(S/C_{V})*V^(-Nk_{B}/C_{V}), where α is a constant.

W_{A-->B}= -αNk_{B}exp(S_{2}/C_{V})∫_{VB}^{VA}V^(-1 - Nk_{B}/C_{V}))dV

= -α exp(S_{2}/C_{V}) C_{V }(V_{B}^(-Nk_{B}/C_{V}) - V_{A}^(-Nk_{B}/C_{V})).

Similarly, W_{C-->D}= α exp(S_{1}/C_{V}) C_{V }(V_{B}^(-Nk_{B}/C_{V}) - V_{A}^(-Nk_{B}/C_{V})).

W_{tot}= α C_{V}((V_{B}^(-Nk_{B}/C_{V}) - V_{A}^(-Nk_{B}/C_{V}))(exp(S_{1}/C_{V}) - exp(S_{2}/C_{V})).

Finding Q_{in}:

Q_{in }= ∫_{S2}^{S1 }TdS = α*V_{B}^(-Nk_{B}/C_{V})∫_{S2}^{S1 }exp(S/C_{V})dS

= α*C_{V}*V_{B}^(-Nk_{B}/C_{V})(exp(S_{1}/C_{V}) - exp(S_{2}/C_{V})).

eff = 1 - (V_{A}/V_{B})^(Nk_{B}/C_{V}).

For a monatomic ideal gas dU = (3/2)Nk_{B}dT = C_{V}dT, Nk_{B}/C_{V}= 2/3, e = 1 - (V_{A}/V_{B})^{-2/3}.

__Refrigerators, Air conditioners, Heat pumps__

A refrigerator uses 10 W of electrical power when it is closed to keep the interior temperature stable. Use reasonable estimates for any relevant temperatures to find an upper bound on the rate at which heat is entering the refrigerator due to imperfect thermal insulation.

Solution:

- Concepts:

The second law of thermodynamics - Reasoning:

For a reversible process Q_{1}/T_{1 }= Q_{2}/T_{2}. This determines the best possible coefficient of performance of the refrigerator. - Details of the
calculation:

For a refrigerator we define the coefficient of performance COP as the ratio of the amount of heat removed at the lower temperature to the work put into the system.

COP = Q_{low}/W = Q_{low}/(Q_{high }- Q_{low}).

The best possible coefficient of performance is

COP_{max}= Q_{low}/(Q_{high }- Q_{low})_{max}= Q_{low}/(Q_{low}(T_{high}/T_{low) }- Q_{low}) = T_{low}/(T_{high }- T_{low}).

Let us assume that T_{low}= 4^{o}C = 277 K, and T_{high}= 20^{o}C. Then COP_{max}= 277/16.

So if the refrigerator were ideal, the heat removed per second = 10W*277/16 = 173 W.

Upper bound on the rate at which heat is entering the refrigerator: 173 W

An electric freezer is turned on inside a tent for a long time. It is 0 ^{
o}C outside the tent, +1 ^{o}C inside the tent, and -13 ^{o}C
inside the freezer. What would be the equilibrium temperature inside the tent
if another freezer is turned on inside the tent? The outside temperature
remains the same. The freezers are identical and follow the Carnot cycle.

Solution:

- Concepts:

Thermal equilibrium, the Carnot cycle - Reasoning:

We have three regions:

Region 1: The inside of the refrigerator

Region 2: The inside of the tent.

Region 3: The outside reservoir at 273 K.

Assume region 1 and region 2 are in equilibrium with their surroundings.

Then the heat flowing into a region per unit time must equal the heat flowing out per unit time. - Details of the calculation:

Region 1: The heat flowing in is proportional to the temperature difference T_{2}– T_{1}.

Q_{1}= b(T_{2}– T_{1})

(ΔQ/Δt = -kAΔT/Δx , where k is the thermal conductivity, ΔT = (T_{2}- T_{1}) is the difference in temperature between side 2 and side 1 of a layer of material of area A, Δx is the thickness of that layer, and ΔQ/Δt is the amount of heat that flows from side 1 to side 2 through the layer of material per unit time. )

This heat Q_{1}is removed again by the refrigerator.

Region 2: The heat flowing in is the heat dumped by the refrigerator.

Q_{2}= Q_{1}T_{2}/T_{1}= b(T_{2}– T_{1})T_{2}/T_{1 }(For the Carnot cycle Q_{1}/T_{1}= Q_{1}/T2.)

The heat flowing out is Q_{3}flowing into region 3 and Q_{1}flowing into region 1. Q_{3}is proportional to the temperature difference T_{2}– T_{3}.

Q_{3}+ Q_{1}= a(T_{2}– T_{3}) + b(T_{2}– T_{1})

In equilibrium Q_{2}= Q_{3}+ Q_{1}.

a(T_{2}– T_{3}) + b(T_{2}– T_{1}) = b(T_{2}– T_{1})T_{2}/T_{1}.

We have a/b = (T_{2}– T_{1})(T_{2}/T_{1}– 1)

With the data for the single refrigerator we can evaluate a/b.

T_{3}= 273 K, T_{2}= 274 K, T_{1}= 260 K, a/b = 0.7538.

For the two refrigerators the wall area is doubled, so b --> b' = 2b, a/b' = 0.3769

T_{3}and T_{1}do not change we have to solve a/b' = (T_{2}– T_{1})(T_{2}/T_{1}– 1) for T_{2}.

T_{2}= 275.43 K = 2.43^{o}C.

A nursery uses natural gas heating to keep the greenhouses at 30 ^{o}C
all year. An engineer points out that the water at the bottom of a nearby lake
is at a constant temperature of 5 ^{o}C, and that he can build an ideal
heat pump that will work at maximum possible efficiency to pump heat from this
lake water into the greenhouses. He claims that the nursery will come out ahead
with his system, even though it uses electricity instead of natural gas at three
times the cost per Joule. Is he right? Neglecting capital and maintenance
costs by what factor would their energy bill change?

Solution:

- Concepts:

The coefficient of performance for a heat pump - Reasoning:

The coefficient of performance for a heat pump is the ratio of the energy delivered at the higher temperature to the work put into the system, COP = Q_{high}/(Q_{high }- Q_{low}). Knowing the COP, we can determine the cost of the delivered energy. - Details of the calculation:

The best possible coefficient of performance is

COP_{max}(heat pump) = (Q_{high}/(Q_{high }- Q_{low}))_{max}= T_{high}/(T_{high }- T_{low}) = T_{room}/(T_{room }- T_{outside})

Here COP_{max}= 303/25.

The gas heating system and the heat pump have to deliver the same amount of energy.

For the heat pump: energy paid for = 25/303 times energy delivered

For the gas heater: energy paid for = energy delivered

Cost for heat pump delivery = 3*25/303 times cost for gas heater delivery

Their new bill would be approximately ¼ their old bill.