1. Thermal Behavior

2. Thermal Properties Heat capacity Specific Heat Thermal Expansion
Thermal Conductivity
Thermal Shock

3. Heat Capacity
As a material absorbs heat, its temperature rises
The HEAT CAPACITY is the amount of heat required to raise its temperature by 1°C
C = Q/∆T
C is the Heat Capacity (J/mol-K),
Q is the amount of heat (J/mol), and
∆T is the change in temperature (K or C)

4. Specific Heat c = q/(m ∆T)
Often we use Specific Heat instead of Heat Capacity because it is per unit mass instead of mol
c = q/(m ∆T)
where c is the specific heat (J/kg-K)
and m is the mass of material being heated (kg)

5. How to measure? There are two ways to measure specific heat:
Volume is maintained constant (and pressure thus builds up), cv
Pressure is maintained constant (and volume increases), cp

6. cv is not a constant

7. But...
The Debye temperature is below room temperature for many solids, so we can still use a value that is useful and approximately constant

8. Material cp (J/kg-K) Aluminum 900 Copper 385 Gold 129 Iron 444 Lead
159
Nickel
Silver
237
Titanium
523
Tungsten
133
Al2O3
160
MgO
457
SiC
344
Carbon (diamond)
519
Carbon (graphite)
711
Nylon 6,6
Phenolic
Polyethylene (high density)
Polypropylene
1880
Polytetraflouroethylene (PTFE)
1050

9. Example
A passive house will include a trombe wall to absorb and store heat. It will be built from 2 kg bricks.
How many bricks are needed to absorb 50 MJ of heat by increasing 10° C?
(The specific heat of the brick is 850 J/kg-K)

10. Example (cont.) q = cp m ∆T => m = q/(c ∆T)
Recall that our target ∆T is 10C = 10K.
m = 50 MJ/(850 J/kgK * 10K) = 5,880 kg
We have 2 kg bricks, so we need
2,940 bricks.

11. Example cont.
Suppose we wanted to accomplish the same goal using water?
The specific heat of water is 1 cal/g-K, with a density of 1 Mg/m3. (1 liter = .001 m3)

12. Example cont
Well, we need to fix our units first. We have been solving things with Joules, but our number is in calories.
There are cals in a Joule. So:
1 (cal/g-K)/ (cal/J) = 4.19 J/g-K

13. Example cont. q = cp m ∆T => m = q/(c ∆T)
Recall that our target ∆T is 10C = 10K.
m = 50 MJ/ (4.19 J/g-K * 10 K) = 1.19 Mg
There are 1,000 grams in a liter of water, so
we need
1,190 liters of water.

14. Example - compare
2,940 bricks at 2 kg each, means 5,980 kg of bricks (almost 6 tons)
1,190 liters of water means 1,190 kg of water (a little over 1 ton)

15. Thermal Expansion

16. Thermal Expansion
An increase in temperature leads to increased thermal vibration of the atoms
This leads to greater seperation distance of the atoms.

17. Thermal Expansion

18. Thermal Expansion ε = α ∆T The percent change in length is given by:
ε = strain
α = coefficient of thermal expansion
∆T = change in temperature

19. CTE vs Temperature

20. Material CTE @ 27C (10-6/C) CTE @ 527C (10-6/C) Fireclay refractory
Aluminum
23.2
33.8
Copper
16.8
20.0
Gold
14.1
16.5
Nickel
12.7
Silver
19.2
23.4
Tungsten
4.5
4.8
Mullite
5.3
Porcelain
6.0
Fireclay refractory
5.5
Al2O3
8.8
MgO
7.6
SiC
4.7
Silica glass
0.5
Soda-lime-silica glass
9.0
Nylon 6,6 30 xx
Phenolic
30-45
Polyethylene (high density)
Polypropylene
68-104
Polytetraflouroethylene (PTFE) 99

21. Strongly bonded solids
General correlations
Weakly bonded solids
Strongly bonded solids
Low melting point
High melting point
Low elastic modulus
High elastic modulus
High CTE
Low CTE

22. Example
We have a tungsten pin that is just a LITTLE too big to fit into the opening in a nickel bar.
The pin is mm in diameter and the hole is mm at room temperature (25 C)
We know that nickel has a higher CTE than tungsten (12.7 x 10-6 vs 4.5 x 10-6), so we figure we can heat them both up and the hole will expand more than the pin and it should fit!
How much should we heat them up?

23. Example continued
Well, we want to heat it up enough so that the diameter of the tungsten equals the diameter of the nickel. If the tungsten diameter increases by ∆dt and the nickel by ∆dn, we want
dt + ∆dt = dn + ∆dn
We know dt = and dn = already, so we want
∆dt = ∆dn

24. Example We know that the change in diameter comes from strain:
∆dt = ε dt
and we know that the strain comes from the temperature change:
ε = α ∆T

25. Example: Plugging together
put all the equations together:
dt + ∆dt = dn + ∆dn
dt + εt dt = dn + εn dn
dt + αt ∆T dt = dn + αn ∆T dn
dt - dn =αn ∆T dn - αt ∆T dt = (αn dn - αt dt)∆T
∆T = (dt - dn)/(αn dn - αt dt)
∆T = ( )/(4.5 x 10-6 x x 10-6 x ) ∆T = 49.4 C = 121 F