Thermo 2nd law

Question 1

what’s the efficiency equation

Answer 1

η = desired output / required input

η = W/Qₕ = (Qₕ - Q꜀) / Qₕ for a heat engine
η = Q꜀/W for a refrigerator

Question 2

what’s the coefficient of performance

Answer 2

the ‘efficiency’ of a refrigerator/heat engine.
not called efficiency cause a refrigerator has it greater than 1

Question 3

Kelvin-Planck statement

Answer 3

it’s impossible to make an engine that exclusively extracts heat from a source and produces an equivalent amount of work.

Question 4

Clausius statement

Answer 4

it’s impossible to make a refrigerator that exclusively transfers heat from a colder to a hotter body

Question 5

Carnot’s theorem

Answer 5

a reversible engine is the most efficient

all reversible engines operating between two heat baths have the same efficiency (η) depending only on T꜀ and Tₕ

Question 6

Carnot efficiency equation

Answer 6

η = 1- Q꜀/Qₕ
= 1- T꜀/Tₕ

Question 7

Clausius inequality

Answer 7

∮1/T đQ ≤ 0
For a system when taken over a cycle

Question 8

entropy equation

Answer 8

dS =đQ/T

(if T is constant then ΔS=ΔQ/T)

Question 9

what’s the change in entropy for a reversible process

Answer 9

the total change in entropy (change in entropy of the universe) is 0
this comes from ΔS≥0. if the forward process has ΔS>0 then the reverse process would have ΔS<0 which isn’t possible

Question 10

how do you find the maximum work extractable when heat is transferred from a finite hot reservoir to an infinite cold one

Answer 10

W= Qₕ - Q꜀
find Qₕ from the equation Qₕ = mcΔT (c: specific heat capacity)

since the cold reservoir is infinite the specific heat is infinite so that equation won’t work. use entropy to find it.
ΔSₕ + ΔS꜀ = 0 (this is the extreme cause which gives the maximum work done)

since T꜀ is constant ΔS꜀= Q꜀/T꜀
ΔSₕ = ∫đQ/T = ∫mc/T dT= mcln(T꜀/Tₕ)
∴ Q꜀ = T꜀mc*ln(T꜀/Tₕ)

then sub in values to get W

Question 11

if heat is transferred from a hot to a cold reservoir and both are finite in size then derive the formula for the final temperature

Answer 11

ΔS = ΔSₕ + ΔS꜀ = 0 (limiting case)
∫đQₕ/T + ∫đQ꜀/T = 0
mcln(T/Tₕ) + mcln(T/T꜀) = 0
mcln(T/Tₕ) = -mcln(T/T꜀)
ln(T/Tₕ) = ln(T꜀/T)
T/Tₕ = T꜀/T
T= sqrt(Tₕ*T꜀)