Thermal Behavior

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

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)

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)

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

cv is not a constant

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

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 1260-2090 Phenolic 1460-1670 Polyethylene (high density) 1920-2300 Polypropylene 1880 Polytetraflouroethylene (PTFE) 1050

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)

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.

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)

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

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.

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)

Thermal Expansion

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

Thermal Expansion

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

CTE vs Temperature

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) 149-301 Polypropylene 68-104 Polytetraflouroethylene (PTFE) 99

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

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?

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

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

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