1. Lecture 22: Temperature and Thermal Expansion
Internal Energy
Temperature
0th Law of Thermodynamics
Thermal Expansion/Contraction 1

2. Internal Energy and Temperature
All objects have “internal energy” (measured in Joules):
random motion of molecules
kinetic energy
collisions of molecules give rise to pressure
Amount of internal energy depends on:
temperature
related to average kinetic energy per molecule
how many molecules
mass
“specific heat”
related to how many different ways a molecule can move
translation
rotation
vibration
the more ways it can move, the higher the specific heat

3. 0th Law of Thermodynamics
If two objects are in thermal equilibrium with a third, then the two are in thermal equilibrium with each other.
If they are in thermal equilibrium, they are at the same temperature.
Perhaps do hot water, and cold, comment on what happens after time.

4. Temperature Fahrenheit Celsius Kelvin 100 373.15 212 Water boils32
Fahrenheit
100
Celsius
273.15
373.15
Kelvin
Water boils
Water freezes
NOTE: K = 0 is “absolute zero”, meaning (almost) zero KE/molecule

5. Thermal Expansion When temperature rises:
molecules have more kinetic energy (moving faster)
consequently, things tend to expand
Amount of expansion depends on…
change in temperature
original length
coefficient of thermal expansion
L =  L0 T (linear expansion)
V =  V0 T (volume expansion)
Temp: T
Temp: T+T L0 L

6. Exception: Water
Water is very unusual in that it has a maximum density at 4 ºC. That is why ice floats (and how we exist)!

7. Summary Temperature Thermal Expansion
Measure of average Kinetic Energy of molecules
Thermal Expansion
L =  L0 T (linear expansion)
V =  L0 T (volume expansion)

8. Example
A brass rivet with diameter cm and a steel plate that has a hole with diameter cm are both at 20ºC. What temperature do they need to cooled or heated to for the rivet to fit in the hole? (brass = 19×10-6 ºC-1 and steel = 12×10-6 ºC-1)
We will use the equation for thermal expansion/contraction:
L =  L0 T

9. Example Ls + Ls = Lb + Lb Ls + s Ls Ts = Lb + b Lb Tb
A brass rivet with diameter cm and a steel plate that has a hole with diameter cm are both at 20ºC. What temperature do they need to cooled or heated to for the rivet to fit in the hole? (brass = 19×10-6 ºC-1 and steel = 12×10-6 ºC-1)
We want the two lengths to be equal:
Ls + Ls = Lb + Lb
Ls + s Ls Ts = Lb + b Lb Tb
Ls + s Ls (Tf-Ti) = Lb + b Lb (Tf-Ti)

10. Ls + s Ls (Tf-Ti) = Lb + b Lb (Tf-Ti)
Example
A brass rivet with diameter cm and a steel plate that has a hole with diameter cm are both at 20ºC. What temperature do they need to cooled or heated to for the rivet to fit in the hole? (brass = 19×10-6 ºC-1 and steel = 12×10-6 ºC-1)
Put in the known values and solve for Tf:
Ls + s Ls (Tf-Ti) = Lb + b Lb (Tf-Ti)
Tf = -51ºC