1. Mechanisms of Heat Flow
If two objects are at different temperatures, energy (heat = Q) will flow between them. For each object,
dQ = CdT  dQ/dt = CdT/dt
Mechanisms of Heat Flow
Conduction
Radiation
Convection
Chigh, Thigh
Clow, Tlow

2. Conduction
Molecules in a material are moving. The higher the temperature, the faster they are moving. They are also interacting (e.g. colliding) with each other, so exchanging energy with their neighbors.
Imagine a temperature gradient in a material. The faster moving molecules at the hot end will lose some of their energy to their slower moving, “colder” neighbors; i.e. energy (heat) will flow from hot to cold.
In a crystal, the molecules are in a periodic array. Instead of considering the motion of individual molecules, consider them to move in compressional waves (i.e. sound waves). Applying the rules of quantum mechanics, quantized sound waves are called phonons.

3. Imagine a slab of material with area A and thickness x, with the hotter side at Th and colder side at Tc. The rate that heat will flow from hot to cold side is:
Power = dQ/dt  A (Th-Tc)/x = A T /x
In differential form: P  A |dT/dx|.
The proportionality constant depends on the material and is called the thermal conductivity (k):
P = kA |dT/dx|
[In general, k is a function of temperature.]

4. Consider a rod of length L and cross-sectional area A:
P = kA dT/dx
kA dT = P dx
kA dT = P dx
If k  constant (e.g. if (Th-Tc) is not too large), kA|Th-Tc| = PL
dQ/dt = P = kA(Th-Tc)/L
[This is an example of “Newton’s Law of Cooling”: the rate of heat flow is proportional to the temperature difference. It is only an approximation (i.e. assumes k  independent of T).] A

5. Since conduction depends on interactions between molecules, gases have lower values of thermal conductivity (since molecules are far apart).
(Generally) metals have higher thermal conductivities than insulators because in addition to molecular interactions, they have electrons which can move around and collide.
[The higher the electrical conductivity of the metal, the more its electrons can move around, so also the higher the thermal conductivity. Notice the low value of k for stainless steel compared to silver – that is why stainless steel cooking utensils don’t feel as hot as silver utensils.]
Stainless steel
Sapphire

6. (Generally) metals have higher thermal conductivities than insulators because in addition to molecular interactions, they have electrons which can move around and collide.
What about diamond? Why is k so large?
1) It is very hard: extremely strong (covalent) bonds between carbon atoms  very large phonon velocities.
2) Phonon motion is not often interrupted by “scattering”: long “mean-free-paths”.

7. Consider an object with heat capacity C at temperature T connected to an infinite heat bath at (constant) temperature T via a thermal conductor (k,A,L).
dQ/dt = CdT/dt
but dQ/dt = (kA/L) (T0-T)
Therefore (CL/kA) dT/dt = T0-T
(CL/kA) dT / (T0-T) = dt
If C and k are  constant (indpt. of T): (CL/kA)  dT /(T-T0) = - dt

8. (CL/kA)  dT /(T-T0) = - dt
Ln(T-T0) = -(kA/CL) t + constant
T = T0 + constant exp(-t/), where the thermal relaxation time
 CL/kA
If T = Ti when t = 0: T = T0 + (Ti-T0) exp(-t/)

9. T = T0 + (Ti-T0) exp(-t/)
fractional change in temperature at any time:
(T-T0) / (Ti-T0) = exp(-t/),
is independent of Ti – T0
(i.e. only depends on t/).

10. C(water) = (4186 J/kgoC) (20 kg) = 83720 J /oC (>> C(beaker)
Problem: What is the thermal relaxation time for a glass beaker holding 20 kg of water. The surface area of the beaker = 0.05 m2 and its walls have thickness = 1 mm. (Assume that the beaker is covered with a perfect thermal insulator.)
C(water) = (4186 J/kgoC) (20 kg) = J /oC (>> C(beaker)
k(glass) = 0.8 W/moC L
 = CL/kA = (83720 J /oC) (10-3 m) / [(0.8 W/moC) (0.05 m2)]
 = 2093 seconds ( 35 minutes)

11. A situation which is also very common but much more difficult to calculate is when one object , at temperature T, is immersed directly into a heat bath at a different temperature, T0. While the surface of the immersed object will quickly reach temperature T0, it will take longer for its interior. The time dependence will depend on the shape of the object, but the characteristic thermal relaxation time will be of the order cr2/(10 k), where c,, and k are the specific heat, density, and thermal conductivity of the object and r is a typical dimension. [The combination D = k/c is called the thermal diffusivity of the material and gives the speed with which heat “diffuses”.]

12. Radiation
All objects emit electromagnetic radiation ; the power emitted:
P = AeT4
where T = temperature (in Kelvin), A = surface area,
 = 5.67 x 10-8 W/m2K4 (= Stefan-Boltzman constant)
and e = “emissivity” is the fraction of the maximum theoretical radiation
that is emitted [0 < e  1 (depends on material)].
Also, if the fraction of incident electromagnetic radiation that is absorbed (absorptivity) = a; can show that e = a.
Therefore, a perfectly black object, which absorbs all incident light has e = 1
(“Blackbody”).
A perfect reflector (e.g. very shiny metal) has e  0 (i.e. e << 1).
[Fingerprints or smudges on a metal will increase e.]

13. P = AeT4
 = 5.67 x 10-8 W/m2K4
Blackbody: e = 1
The spectrum of the emitted radiation from a blackbody also depends on temperature. The peak in the spectrum occurs at a wavelength:
peak = 2.9 x 10-3 mK/ T
Blackbody Spectrum
T(K)
P/A (W/m2)
peak 30
0.046
0.10 mm (FIR)
100
5.67
29 m (FIR)
300
459
10 m (mid-IR)
1000
5670
2.9 m (mid-IR
3000 (bulb)
4.59 x 106
1 m (NIR)
5780
(sun surface)
6.3 x 107
0.5 m (cyan)
40,000K
(blue giant)
1.5 x 1011
73 nm (uv)

14. Suppose an area A of a cold blackbody (at TC) is facing an area A of a hot blackbody (at TH).
The hot object emits power P = ATH4 which the cold object absorbs.
The cold object emits power P = ATC4 which the hot object absorbs.
Therefore, the rate of heat flow (in the form of EM radiation):
P = dQ/dt = A(TH4 –TC4)
If the emissivity e<1:
P = dQ/dt = Ae(TH4-TC4) TC TH
must use Kelvin !!!

15. Dewar (Thermos) for storing hot or cold liquids:
Vacuum between inside and outside
(so dQ/dT(cond)  dQ/dT(convection)  0)
Silver both the inner and outer surfaces of the vacuum
(so e <<1 to minimize dQ/dT(rad)
Remaining heat leaks:
a) radiation: e  0
b) conduction through walls supporting the
inner chamber.
c) conduction, convection, radiation through
opening.

16. Problem: 0.5 kg of ice at 0oC is inside a thermos with surface areas 0.2 m2 of the inner and outer wall. How long will the ice last (before completely melting) if the emissivity of the walls e = 0.05? (Assume the outer wall is at room temperature = 295 K and that all the heat leak is via radiation.)

17. The rate of heat flow (power) = dQ/dt = Ae(TH4 – TC4)
Problem: 0.5 kg of ice at 0oC is inside a thermos with surface areas 0.2 m2 of the inner and outer wall. How long will the ice last (before completely melting) if the emissivity of the walls e = 0.05? (Assume the outer wall is at room temperature = 295 K.)
To melt all the ice, must transfer Q = mLf = (0.5 kg) (3.33 x 105 J/kg) = 1.67 x 105 J.
The rate of heat flow (power) = dQ/dt = Ae(TH4 – TC4)
= (5.67 x 10-8 W/m2K4)(0.2 m2) (0.05) ( )K4
= 1.14 W
 t = Q/(dQ/dt) = 1.67 x 105 J/ 1.14 J/s = 1.46 x105s = 4.1 hours
[Note: if the thermos was not “silvered” well, so e  1,
the ice would only last 0.2 hours = 12 minutes.]

18. dQ/dt = Ae(TH4-TC4) (exact)
If TH = TC + T, dQ/dT = A e[(TC+T)4 – TC4]
If T << TC, expand to first order in T :
dQ/dT = A e[(TC4 + 4TC3T + …) – TC4]
dQ/dT  4AeTC3 T (approx.)
[This is another example of Newton’s Law of Cooling and Heating, but only valid if TH - TC << TC] TC TH

19. Problem: Compare the rates of heat flow due to conduction and due to radiation between blackbodies at TH and TC (= 300 K), separated by a distance L = 1cm of air.
(Assume TH-TC << TC) [kair = W/mK]

20. Problem: Compare the rate of cooling due to conduction and due to radiation of
blackbodies at TH and TC (= 300 K), separated by a distance L = 1 cm.
(Assume TH-TC << TC) [kair = W/mK]
dQ/dT (conduction) = kair TA/L
dQ/dT(radiation)  4ATC3 T
dQ/dT(radiation) / dQ/dT(conduction) = 4TC3L/kair
dQ/dT(rad) / dQ/dT(cond) = 4 (5.67 x10-8 W/m2K4) (300 K)3 (0.01 m) / W/mK
dQ/dT(rad) / dQ/dT(cond) = 2.7
[Would need L < (1 / 2.7 cm) = 3.7 mm for conduction to be greater than radiation. However, if this was a silvered thermos with e = 0.05, conduction would be greater than radiation if L< 3.7mm/0.05 = 74 mm: “soft dewar”]