MECHANICAL ENGINEERING: Thermodynamics - Entropy and the Second Law of Thermodynamics

Entropy

- a measure of a system's energy that is unavailable for work

- a measure of the disorder or randomness

- from Greek entropia "a turning towards"



Second law of thermodynamics

- there is no machine with 100% efficiency

- phenomenon of irreversibility in nature

- In a system, a process that occurs will tend to increase the total entropy of the universe

- The entropy of an isolated system which is not in equilibrium will tend to increase over time, approaching a maximum value at equilibrium

- It is impossible for heat to flow from a colder body to a warmer body without any work having been done to accomplish this flow

- It is impossible to extract an amount of heat from a hot reservoir and use it all to do work. Some amount of heat must be exhausted to a cold reservoir



Few examples of Entropy and the 2nd Law of Thermodynamics

- flow of heat from a hot object to a cold object

- flow of a gas from high pressure to low pressure

- evaporation of water

- melting of ice

- cloud formation in the sky

- adding sugar to a cup of coffee

- carbon dioxide is dissolved in water

- spontaneous (natural) cleaning of a messy house

- shuffling of playing cards

- breaking a mirror or glass



1. Isothermal and reversible process


dS = Q/T


where:

dS = change in entropy, values can be (+)increase or (-)decrease

Q = heat, negative (-) if heat removal, positive (+) if heat addition

T = absolute temperature




2. Temperature not constant


dS = Integral (dQ/T)


dQ = m * c * dT


substituting,

dS = Integral (dQ/T)

dS = Integral (m * c * dT/T)

dS = m * c * Integral (dT/T)


if

T1 = initial temperature

T2 = final temperature


dS = m * c * Integral(T1,T2) (dT/T)


then


dS = m * c * ln T2/T1


where:

m = mass

c = specific heat

dS = change in entropy, values can be (+)increase or (-)decrease

dQ = heat, negative (-) if heat removal, positive (+) if heat addition

T = absolute temperature




1. Find the change in entropy if 1 kg of water at 20 C is changed into ice at 0 C.


find:

dS = total change in entropy


given:

m = 1 kg of water

c of water = 4.2 kJ/kg K

lf = heat of fusion of ice = 334 kJ/kg

T1 = 20 C + 273 = 293 K

T2 = 0 C + 273 = 273 K


solution:

dS = dS1 + dS2


dS1 --> cooling of water


dS1 = m * c * ln T2/T1

dS1 = 1 * 4.2 * ln 273/293

dS1 = - 0.3 kJ/K


dS2 --> freezing of water to ice

dS2 = Q/T


but

Q = - m * lf  --> latent heat


note: Q is negative (-) because of HEAT REMOVAL


substituting,

dS2 = - m * lf/T

dS2 = - 1 * 334/273

dS2 = - 1.22 kj/K


total change in entropy

dS = dS1 + dS2

dS = (- 0.3) + (- 1.22)

dS = - 1.52 kJ/K




2. Calculate the change in entropy when 4 kJ of heat flows from the warm exterior at 40 C into a house at 22 C.


find:

dS = total change in entropy


given:

Q = 4 kJ

T exterior = 40 C + 273 = 313 K

T house = 22 C + 273 = 295 K


solution:

dS = dS exterior + dS house


dS exterior:


The entropy of the exterior decreases, since heat flows out of it (- Q)

dS exterior = -Q/T exterior

dS exterior = -4/313

dS exterior = -0.013 kJ/K



dS house:

The entropy of the house increases since heat flows into it (+ Q)

dS house = Q/T house

dS house = 4/295

dS house = 0.014 kJ/K



total change in entropy

dS = dS exterior + dS house

ds = -0.013 + 0.014

dS = 0.001 kJ/K