Thermal Behavior

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 • 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

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 0.2389 cals in a Joule. So: • 1 (cal/g-K)/ 0.2389 (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 • The percent change in length is given by: • ε = α ∆T • ε = strain • α = coefficient of thermal expansion • ∆T = change in temperature

19. CTE vs Temperature

21. General correlations

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 5.000 mm in diameter and the hole is 4.999 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 = 5.000 and dn = 4.999 already, so we want • 5.000 + ∆dt = 4.999 + ∆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 = (5 - 4.999)/(4.5 x 10-6 x 4.999 - 12.7 x 10-6 x 5.000) ∆T = 49.4 C = 121 F