Thermodynamics Ch 16 – Temperature

Temperature and the Zeroth Law of Thermodynamics

The Zeroth Law of Thermodynamics

If two objects are separately in thermal equilibrium with a 3^rd object, then the first two objects are in thermal equilibrium with each other.

Thermal equilibrium: If two objects are in thermal equilibrium then neither object will exchange energy by heat or electromagnetic radiation if they are placed in thermal contact.

Temperature is a property that determines whether an object is in thermal equilibrium.

Thermal contact: If in thermal contact two objects will exchange energy between them if a temperature difference.

Thermometers and the Celsius Temperature Scale

You shouldn’t ever have to memorize formulas in Physics. Take for example the formula from or to Celsius to/from Fahrenheit.

WATER	Freezing	Boiling	Difference
Fahrenheit	32°	212°	180°
Celsius	0°	100°	100°

Liquid Nitrogen Demo: TH-A-LN

Take the ratio of

180/100 = 9/5 & 100 / 180 = 5/9

Where is the “zero” (with respect to water) for the Celsius scale?

Ans: 0 °C

For the Fahrenheit scale?

Ans:32 °F

Now, which scale has the most number between freezing and boiling?

Fahrenheit (we’ll call this the “BIG” scale; conversely Celsius is the “small” scale)

So we say the “zero” point for the Fahrenheit scale is 32 °F…

· I think most people will agree this is a weird “zero” point

· How can we bring this “zero” point to zero?

o Ans: Subtract 32 °F

· We also recognize that the Fahrenheit scale is the “BIG” scale…so how can we bring it to a small scale?

o Ans: Multiply by a small ratio, 5/9

Let’s try using this logic…

98.6 °F

Let’s bring the Fahrenheit scale down to a more logical “zero” point.

· 98.6 °F - 32 °F = 66.6 F°

à (take note of the units, 66.6 F°)

We now need to bring this “BIG” scale down to the “small” scale by multiplying by 5/9

· 66.6 F° * 5/9 = 37 °C

Let’s go in reverse…

37 °C uses the “small” scale

à let’s go to the “BIG” scale

· 37 °C * 9/5 = 66.6 F°

But the “zero” point of the Fahrenheit scale isn’t “zero”…it’s 32 °F

· 66.6 F° + 32 F° = 98.6 °F

Thermometers make use of thermometric properties.

Types of thermometers

· volume expansion (Hg column)

· constant volume gas (Δ gas pressure)

· thermocouples (Voltage differences of two different metals)

· electrical resistance (Platinum wire)

· thermographic (emitted radiation)

Constant-Volume Gas Thermometer and

the Absolute Temperature Scale

The Kelvin scale is the proper scale in most situations (PV = nRT, etc). When in doubt…just use the Kelvin scale. Why?

1 Kelvin = 1 °C and T = T_C + 273.15

So if the temps are given in °C then ΔT will be the same whether in the Kelvin or Celsius scales.

If you notice

T à temperature in the Kelvin scale

T_C à temperature in Celsius

T_F à temperature in Fahrenheit

The Kelvin scale is an absolute scale which means you can not go below zero.

At 0 K (do not use the degree sign) all movement (i.e. in an atom) stops. As you can see, we can not achieve the Absolute Zero point, but we can achieve temps very close.

Demo: Galileo’s Thermometer: TH-E-GT

Demo: Absolute Zero: TH-E-AZ

Thermal Expansion of Solids and Liquids

Let’s pretend we have a 10 cm steel ruler an a 2 meter stick (also steel)

If we heat both of the steel sticks…which will expand more a greater amount?

Most of you will say the long one will expand more and you’ll be RIGHT!

Some may say they will expand the same and you’ll be CORRECT also (almost)!

Explanation:

Since they are both steel, they will expand by the same proportion; but not the same amount…

(We know the longer one will expand by the greater total length since 5% of 2 meters is much greater than 5% of 10 centimeters)

Now let’s pretend we have two 10 cm steel rulers, A and B.

10 cm steel ruler, A, undergoes ΔT_C = 10 C°

10 cm steel ruler, B, undergoes ΔT_C = 100 C°

Which will expand more?

B, will expand a factor of 10 greater A

Most of you already knew these results.

These results also lead to one conclusion:

ΔL = α L ΔT

Which states that the change of length of an object is dependant on

· Type of material

· Initial Length of material

· Change of temperature

Application of Thermal Expansion/Contraction

Question: What would happen if the coupling and the pipes were made of different materials and the entire new system became very hot or very cold?

Ans: Different materials have different coefficients of linear expansion, so if the coupling is a different material, the coupling would either crack (contract too much) or leak (expand too much)

This is the same principle as the bimetallic strip discussed in your book.

You are attempting to form the best junction between two pipes.

Use a coupling between pipes

· heat the coupling (torch)

· cool the ends of the pipes (ice bath or oil bath cooled with liquid nitrogen)

What happens to the coupling?

· it expands, including the hole in the middle

What happens to the pipes?

· they contract, become smaller

Slide the pipes into the coupling…let achieve room temp.

You now have a coupling as strong as the pipe itself.

Volume Thermal Expansion

Derivation:

à Remember Young’s Modulus…we now have another application

Young’s Modulus = Stress / Strain; where stress F/A and strain is ΔL/L

(Y) = F/A / ΔL/L

(Y) ΔL/L = F/A (= stress)

Stress = (Y) ΔL / L

; where ΔL = α L ΔT

Stress = (Y) α L ΔT / L

Stress = Y α ΔT

So if you know Young’s Modulus and the coef of linear expansion, you can calculate Stress.

First NOTE:

β equals 3 * α

Coef of Volume Expansion

= 3*(Coef of Linear Exp)

· Derivation below

And the formula follows similarly to linear expansion ΔL = α L ΔT

ΔV = β V ΔT

One of the notable exceptions to materials contracting as they get colder.

WATER!!! Water’s highest density is at 4 °C

This is very important for life, if you don’t see why…ask in CLASS!!!

V_o + DV = (l + Dl) (h + Dh) (w + Dw)

V_o + DV = (l + alDT) (h + ah DT) (w + aw DT)

V_o + DV = l*h*w (1 + aDT)³

V_o + DV = V_o (1 + 3aDT + 3(aDT)² + (aDT)³)

DV/V_o = 3aDT + 3(aDT)² + (aDT)³

We also know aDT <<1 for most materials…so

DV/V_o = 3aDT

DV = V_o3aDT

Demo: Thermal Expansion of Solids: TH-A-TS

Demo: Thermostat: TH-A-DT

Macroscopic Description of an Ideal Gas

PV = nRT à R = 8.314 J/(mole*K) and n = m/M where m is mass of material

PV = NRT / N_A

Let k_B = R / N_A

PV = N k_BT

n = N / N_A

N_A = 6.022x10²³ molecules/mole

N is number of molecules

k_B = 1.38 x 10^-23 J/K

M = molar mass of substance

(from Periodic Table)

m_H = 1 amu

M_H = 1 gram/mole

(2 protons & 2 neutrons)

m_He = 4 amu

M_He = 4 grams/mole

Some students have confusion between Molar Mass and Molarity

http://www.ilpi.com/msds/ref/concentration.html#molarity

Molarity. Molarity is the number of moles of solute dissolved in one liter of solution. For example, if we have 90 grams of glucose (molar mass = 180 grams per mole) this is (90 g)/(180 g/mol) = 0.50 moles of glucose. If we place this in a flask and add water until the total volume = 1 liter we would have a 0.5 molar solution. Molarity is usually denoted with an italicized capital M, i.e. a 0.50 M solution.

Recognize that molarity is moles of solute per liter of solution, not per liter of solvent!! Also recognize that molarity changes slightly with temperature because the volume of a solution changes with temperature.