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Thermodynamics

The study of heat energythrough random systems

Work done ON a gas

Comes from

Work done by a gas

Laws of Thermodynamics0th Law: If Ta=Tb and Tb=Tc then

Ta=Tc

1st Law: ΔU=Uf - Ui=W+Q (Conservation of Energy)The increase in internal energy of a

system is determined by the heat of the system and the work added to it.

The idea behind the Heat Engine

VocabΔU=W+Q W=-P V

Isovolumetric- Constant Volume=nC v

Isothermal- Constant temperature

Isobaric- Constant PressureQ

Adiabatic- No Heat Exchange (also no change in entropy)

Q= 0 & ΔS=0 =constant

A PV diagram can be used to represent the state changes of a system, provided the system is always near equilibrium.

1st Law & Thermodynamic ProcessesProcess ΔU Q W

Isobarric nCvΔT nCpΔT - PΔV

Adiabatic nCvΔT 0 ΔU

Isovolumetric nCvΔT ΔU 0

Isothermal 0 - W

GeneralnCvΔT ΔU - W

(PV Area)U=

Table 12.2 for Monatomic Ideal Gas

2nd Law: A system can be defined by its entropy. In a closed system entropy tends to increase.Entropy: A measure of the disorder

(randomness) of the system3rd Law: Temperature and Entropy

are absolute scales.At some point no temperature or

entropy exists.T→0 K and S→S0

Heat Engines Engines allow heat energy to be

transformed into work or mechanical energy.Work or Energy Heat increases

no big deal. Friction does this.Heat Work or Energy

is a big deal. Heat is easy to move around. You could just bring heat wherever you needed work done and “Boom!” you wouldn’t have to do the work, a machine could.

Work or Energy Heat decreases is also a big deal. Making food cold preserves it and

allows it to be moved readily. Less spoilage means less disease.

Any heat engine works on the same properties. A hot reservoir is the source of the

energy. Both words mean something. Hot means

that there is plenty of heat energy and reservoir means that if heat is removed the temperature doesn’t drop much.

There is also a need for a cold reservoir. Again, both words mean something. Cold because it is at a lower temperature

than the hot reservoir and reservoir because it must be large enough that you can dump heat into it without appreciably raising the temperature.

What happens if we put a hot and cold reservoir in contact? Thermal Equilibrium is

not the answer! Heat transfer (or flow) is the answer.

Remember that these are reservoirs so it would take a long time for them to come into thermal equilibrium.

This is great, but we don’t get any work out of it.

We need to “steal” some of the energy leaving the hot reservoir and make it do work for us.

Heat Engine

• Stealing some energy to do work

High Temp

Low Temp

ΔE

Q

W

Types of Heat EngineSteam Engine

Internal Combustion Engine

Unfortunately we don’t get an even trade!

We lose energy to randomness/EntropyThis is the 2nd Law of Thermodynamics

Automobile engines are only about 15% efficient. That means for every 100J of heat energy, 15J worth of work is done on the piston and 85J of heat are discarded. Still, this is the source of energy for most of our transportation.

h

c

h

ch

h

net

QQ

QQQ

QWeff

1

The efficiency of an engine is defined by

𝑒𝑓𝑓=𝑁𝑒𝑡𝑊𝑜𝑟𝑘𝑑𝑜𝑛𝑒𝑏𝑦 h𝑡 𝑒𝑒𝑛𝑔𝑖𝑛𝑒𝐻𝑒𝑎𝑡 𝑖𝑛𝑝𝑢𝑡

If a steam engine takes in 2.254 x 104 kJ of heat and gives up 1.915 x 104 kJ of heat to the exhaust, what is the engines efficiency? How much work is done?

h

c

QQ

1eff

15%or 10 x 2.25410 x 1.915

4

4

15.

1

kJkJeff

Carnot Engine Maximum efficiency

In theory could run backwards

All temperatures in Kelvin What is the efficiency of an ideal steam

engine with steam at 685 K and exhaust at 298 K?What is Qhot if Qcold is 450J?

h

ch

TTT

inputHeatoutputWorkeff

a.

b.

%5656.685

298685 or

K

KKeff

hh

c

Q450J

QQ

156.1eff

J1022)156(.

450JQh

We can go backwards!

• This is a refrigerator or air conditioner

High Temp

Low Temp

ΔE

Q

W

This requires Energy/WorkThis is why your refrigerator must be

plugged in.It is constantly dumping heat into your

kitchenDue to the 2nd Law of

Thermodynamics more heat is dumped than is removedIf you left the refrigerator door open you

would heat up the kitchen

Perfect Heat Engine

Qcold

Qhot

ΔE

W

Qcold

Qhot

ΔE

W

EfficiencyWnet=Qnet=Qhot-Qcold

or Eff=Wnet/Qhot=(1-Qcold/Qhot)

Ex: A steam engine absorbs 1.98 x105J and expels 1.49 x105J in each cycle. Assume that all the remaining energy is used to do work.a. What is the engine’s efficiency?b. How much work is done in each

cycle?

a. Eff=0.247b. W=4.9 x104J

More 2nd Law Entropy

S=entropyQ=Heat (Joules)T=Temperature (In Kelvin)

Entropy in a system must increase or at least stay the same!!!!!!!!!!!!

TQS

An engine has a hot reservoir at 1000 K and uses the atmosphere at 300 K as the cold reservoir. You take 2500 J from the hot reservoir to do 1900 J of work.A. How much heat goes into the

atmosphere?B. Is this engine possible? (Does the

entropy increase?)

Wnet=Qnet=Qhot-QcoldQcold=Qhot-Wnet=2500J -1900J =600J

Entropy

Engine is not possible. What is the maximum amount of work we can

take out? How much work is done?

KJKJ

TQShot /5.2

10002500

KJKJ

TQScold /0.2

300600