Thermal BehaviorThermal Behavior

Thermal PropertiesThermal Properties

• Heat capacity

• Specific Heat

• Thermal Expansion

• Thermal Conductivity

• Thermal Shock

• Heat capacity

• Specific Heat

• Thermal Expansion

• Thermal Conductivity

• Thermal Shock

Heat CapacityHeat 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)

• 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 HeatSpecific 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)

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?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

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 constantcv is not a constant

But...But...

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

• 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 444

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

ExampleExample

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)

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.)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.

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.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)

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 contExample 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

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.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.

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 - compareExample - 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)

• 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 ExpansionThermal Expansion

QuickTime™ and a decompressor

are needed to see this picture.

Thermal ExpansionThermal Expansion

• An increase in temperature leads to increased thermal vibration of the atoms

• This leads to greater seperation distance of the atoms.

• An increase in temperature leads to increased thermal vibration of the atoms

• This leads to greater seperation distance of the atoms.

Thermal ExpansionThermal Expansion

Thermal ExpansionThermal Expansion

The percent change in length is given by:

ε = α ∆Tε = strain

α = coefficient of thermal expansion

∆T = change in temperature

The percent change in length is given by:

ε = α ∆Tε = strain

α = coefficient of thermal expansion

∆T = change in temperature

CTE vs TemperatureCTE vs Temperature

Material CTE @ 27C (10-6/C) CTE @ 527C (10-6/C)Aluminum 23.2 33.8

Copper 16.8 20.0

Gold 14.1 16.5

Nickel 12.7 16.8

Silver 19.2 23.4

Tungsten 4.5 4.8

Mullite 5.3 5.3

Porcelain 6.0 6.0

Fireclay refractory 5.5 5.5

Al2O3 8.8 8.8

MgO 7.6 7.6

SiC 4.7 4.7

Silica glass 0.5 0.5

Soda-lime-silica glass 9.0 9.0

Nylon 6,6 30 xx

Phenolic 30-45 xx

Polyethylene (high density)

149-301 xx

Polypropylene 68-104 xx

Polytetraflouroethylene (PTFE)

99 xx

General correlationsGeneral correlations

Weakly bonded solids

Strongly bonded solids

Low melting point High melting point

Low elastic modulus

High elastic modulus

High CTE Low CTE

ExampleExample

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?

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 continuedExample 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

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 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

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

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

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