Thermodynamics

What is Heat?We use the term heat, but so far we have only defined temperature.

Heat is a type of energy. When you add heat to something you tend to increase its temperature. When you remove heat from something you tend to decrease its temperature.

Chemists (and others) use the unit of calorie for heat. A calorie can be converted to our MKS unit of energy the joule by

4.186J = 1cal

Internal EnergyThe internal energy of a substance is the sum of all of the energy of all of the molecules in a substance.

This is different then the temperature, the temperature is the average kinetic energy of each molecule in a substance.

Something with only 10 molecules could have a very high temperature. Something with a lower temperature but vastly more molecules is likely to have a greater internal energy.

Heat is the flow of energy from one substance to another. Note that the direction of heat flow depends on temperature, not the internal energy.

Specific HeatThe rate at which heat is converted into temperature in a substance is called the specific heat and its value depends on the substance. The following equation shows this relationship

where Q is the heat energy, m is the objects mass and c is the specific heat.

For water at 15℃ and 1 atm of pressure it takes 1kcal to raise 1 kg of water 1 degree. That is the specific heat of water is

Q = mc�T

c = 1kcal/(kg · C) = 4186J/(kg · C)

Thermal Equilibrium

If two substances are brought into contact heat will flow from the hotter one to the colder one until the two have the same temperature.

The time this takes to occur depends on the substance, but given enough time the same temperature will be reached.

Since heat is just energy, energy conservation tells us that the heat that flows from one substance is equal to the heat gained by the other substance.

QL = QG

Example 19-3If 200cm3 of tea at 95℃ is poured into a 150g glass cup initially at 25℃ what will be the common final temperature T when equilibrium is reached, assuming no heat flows to the surroundings?

knownV = 200cm3

Tt = 95C

Tc = 25C

unknownT = ? physics

QL = QG

Q = mc�T

tea is mostly waterctea = 4186J/kg ccup = 840J/kg

mtea = ⇢H2OVtea = (1000kg/m3)(200⇥ 10�6m3) = 0.20kg

Qtea = Qcup = mteactea(95� T ) = mcupccup(T � 25)

mcup = 0.15kg

(0.2kg)(4186J/kg)(95� T ) = (0.15kg)(840J/kg)(T � 25)

79, 500J � (837J/C)T = (126J/C)T � 3150J

T = 86�C

Latent HeatWhen ice melts or water boils the water changes state with out changing temperature. However, there is still heat absorbed in these transitions.

This heat is called latent heat. Instead of increasing the temperature of the substance the energy is used to change state. The formula for this is very straightforward it is just

where L is the latent heat which depends on the substance. Notice there is no temperature dependence. This heat does not change the temperature, it just changes the state. The latent heat is called the heat of fusion when converting solids to liquids and the heat of vaporization when converting liquids to solids.

Q = mL

Example 19-5A 0.50kg chunk of ice at -10℃ is placed in 3.0kg of ‘iced’ tea at 20℃. At what temperature and in what phase will the final mixture be? The tea can be considered as water. Ignore any heat flow to the surroundings.

knownmice = 0.50kgTice = -10℃mtea = 3.0kgTtea = 20℃

unknownT = ?

physics

First we have to figure out what the final state will be, all liquid, all solid, or a mixture.

To bring the water to 0℃ takes: Q = mc�T = (3.0kg)(4186J/kg)(20C) = 250kJ

To bring the ice to 0℃ takes:To melt the ice takes:

250 kJ > 177.5 kJ so we will be left with all liquid ‘iced’ tea

Q = mc�T = (0.5kg)(2100J/kg)(�10C) = �10.5kJ

Q = mL = (0.5kg)(333kJ/kg) = �167kJ

Example 19-5

So the heat gained by the ice is the amount to heat it to 0℃, the amount to melt it, and then the amount to raise it to the equilibrium

temperature T. This is the same amount as the heat lost by the ‘iced’ tea.

= micec(T � 0) + 167kJ + 10.5kJQtea = mteac(20C � T )

(3.0kg)(4186J/kg)(20C � T ) = (0.5kg)(4186J/kg)(T � 0) + 177.5kJ

T = 5.0�C

A 0.50kg chunk of ice at -10℃ is placed in 3.0kg of ‘iced’ tea at 20℃. At what temperature and in what phase will the final mixture be? The tea can be considered as water. Ignore any heat flow to the surroundings.

Return of the Jedi - Star Wars Episode VI

How much energy would it take to unfreeze Han Solo?

which is a whole lot of energy. In terms of kinetic energy it would be equivalent of changing Han Solo's velocity by about 82m/s.

  Since the human body is mostly water we can just pretend a person is water, but at temperatures below 0ºC water is frozen as ice.  So to calculate the heat needed for this

temperature change we need the specific heat of water, cw = 4186 J/kgºC, the specific heat of ice, ci = 2093 J/kgºC, and the latent heat of fusion for water, Lf = 333 kJ/kg. Given

these facts we can write the heat energy as:

Let's assume Han Solo has a mass of 80kg.

Carbonite is apparently a form of frozen carbon dioxide which has a melting point of -56.6ºC.

Human body temperature is 37ºC so Han's body temperature has to be raised ΔT=(37 - -56.6) = 93.6ºC.

You can tell Han Solo is very cold from his shaking so he might not be all the way to 37ºC; however, he must be very close. You get hypothermia at only 2ºC lowering of your core temperature.

Q = m(ci�Ti + cw�Tw + Lf ) =

180kg(2093J/(kgC)(56.6) + 4186J/(kgC)(37) + 333, 000J/kg) = 606.3kJ

First Law of Thermodynamics

The first law of thermodynamics is that the change of internal energy of a system is equal to the heat added to the system minus any work the system does to its surroundings. That is

This is just another way of saying energy is conserved. It has a separate name only because these ideas were developed by different people in different places over many years and only much later did we realize they were saying the same thing. Including all the forms of energy we have discussed we can write

�Eint = Q�W

�K +�U +�Eint = Q�W

Example 19-3A 3.0g bullet traveling at a speed of 400m/s enters a tree and exists the other side is with a speed of 200m/s. Where did the bullet’s lost kinetic energy go and what was the energy transfered?

known

m = 0.003kgvi = 400m/svf = 200m/s

unknown

ΔK = ?

physics

energy conservation

Q = W = ΔU = 0

The kinetic energy lost goes into internal energy of the tree.

�Eint = ��K = �1

2m(v2f � v2i ) = �1

2(0.003kg)

⇥(200m/s)2 � (400m/s)2

⇤= 180J

Example 3.3Polishing a Fitting - A machinist polishes a 0.50-kg copper fitting with a piece of emery cloth for 2.0 min. He moves the cloth across the fitting at a constant speed of 1.0 m/s by applying a force of 20 N, tangent to the surface of the fitting. (a) What is the total work done on the fitting by the machinist? (b) What is the increase in the internal energy of the fitting? Assume that the change in the internal energy of the cloth is negligible and that no heat is exchanged between the fitting and its environment. (c) What is the increase in the temperature of the fitting?

knownm = 0.5 kg

Δt =2 min = 120sv = 1.0 m/sF = 20 N

unknownW = ?ΔEint = ?ΔΤ = ?

PhysicsP = F. v

ΔEint = Q - W

Q = mcΔΤ

P =dE

dt=

�E

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! �E = P�t = Fv�t = (20N)(1.0m/s)(120s) = 2.4⇥ 103J<latexit sha1_base64="mIzOnsGddPkTGaiKxPilMwd5fac=">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</latexit><latexit sha1_base64="mIzOnsGddPkTGaiKxPilMwd5fac=">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</latexit><latexit sha1_base64="mIzOnsGddPkTGaiKxPilMwd5fac=">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</latexit><latexit sha1_base64="mIzOnsGddPkTGaiKxPilMwd5fac=">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</latexit>

�Eint = 2.4⇥ 103J<latexit sha1_base64="V6R0KKMqkXGAm78kzb5Gka+LU/I=">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</latexit><latexit sha1_base64="V6R0KKMqkXGAm78kzb5Gka+LU/I=">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</latexit><latexit sha1_base64="V6R0KKMqkXGAm78kzb5Gka+LU/I=">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</latexit><latexit sha1_base64="V6R0KKMqkXGAm78kzb5Gka+LU/I=">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</latexit> �T =

Q

mc=

2.4⇥ 103J

(0.5kg)(3.9⇥ 102J/kg�C)= 12�C

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Heat TransferHeat can be transferred between places by three different methods.

Conduction - is heat transfer by molecules that are touching one another.

Convection - is heat transfer by hot and cold material moving in bulk to different locations.

Radiation - is heat transfer from electromagnetic waves.

Second Law of Thermodynamics

We have learned that when heat is transferred between objects energy is conserved. But energy conservation allows heat to flow from hot to cold or from cold to hot.

The second law of thermodynamics states that heat will only flow spontaneously from a hot to a cold object; never from a cold object to a hot object.

Another way to express the second law of thermodynamics is to say that entropy must always increase.

EntropyEntropy is a measure of the disorder of a system. The more disordered the system the greater the entropy. Entropy can be expressed as

The second law of thermodynamics states that when heat flows the entropy or disorder of the system increases. Note this equation is only good if the temperature doesn’t change.

�S =Q

T

Forward by Messe Kopp

Example 4.4Entropy Change of Melting Ice - Heat is slowly added to a 50-g chunk of ice at 0°C until it completely melts into water at the same temperature. What is the entropy change of the ice?

knownm = 50g = 0.050 kg

T = 0ºC = 273K

unknown

ΔS = ?

physics

ΔS = Q/T

Q = mL

Lice = 333kJ/kg

ΔS = mL/T = (0.05kg)(333kJ/kg)/273K = 61.5 J/K

Entropy with Changing TIf the temperature changes it doesn’t mean the entropy doesn’t change, it just means we have to integrate to calculate it.

In general this is complicated to solve which is why we don’t do it here. But remember our simple formula is only good for constant T.

dS =dQ

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Homework

Vol 2, Chapter 3 - 22, 26

Vol 2, Chapter 4 - 44, 46