Rate of heat flow

Unit 11 asked how much heat. This unit asks how fast. Heat travels from a hot place to a cold one by three mechanisms — conduction, convection and radiation — and each has its own law. Conduction passes energy along without moving the material, convection carries it bodily in a moving fluid, and radiation sends it across empty space as electromagnetic waves. A thermos flask is the standard exam object because it is designed to defeat all three at once.

12.1 Conduction: Thermal conductivity and measurement

Hold one end of an iron rod in a fire and the other end soon becomes too hot to hold. The atoms at the hot end vibrate more violently, collide with their neighbours and pass the energy along the lattice; in a metal the free electrons carry most of it, which is why metals conduct far better than non-metals. No part of the rod moves as a whole.

It is often convenient to write the law like Ohm's law. Defining the thermal resistance R = L/(kA),

so that temperature difference plays the part of potential difference and heat current the part of electric current. Slabs one behind the other are in series, and their thermal resistances simply add: R = R_1 + R_2 + \cdots. This single idea handles every composite-wall problem.

Material	k / W m⁻¹ K⁻¹	Material	k / W m⁻¹ K⁻¹
Silver	420	Glass	0.8
Copper	390	Brick	0.6
Aluminium	240	Water	0.6
Brass	110	Wood	0.15
Steel	50	Air (still)	0.024

Measuring k of a good conductor: Searle's method

A thick lagged bar of the metal is heated at one end by a steam chest and cooled at the other by water flowing steadily through a spiral. Two thermometers \theta_1 and \theta_2 sit in holes a measured distance x apart along the bar; two more read the inlet and outlet temperatures \theta_3 and \theta_4 of the water.

For a bad conductor such as glass, cardboard or rubber, Searle's bar is useless: the flow would be far too small to measure. Lee's disc method is used instead. The specimen is made a thin disc of large area sandwiched between a steam chamber and a thick brass disc, so that the heat flows through a short distance and a wide area.

12.2 Convection

Warm a fluid from below and it expands, becomes less dense, and is pushed up by the denser cold fluid around it, which then takes its place and is warmed in turn. The resulting circulation is a convection current. Because the material itself has to move, convection is impossible in solids and impossible in a vacuum.

	Natural convection	Forced convection
Cause of motion	density difference and gravity	pump, fan or blower
Speed	slow	fast
Example	air above a heater, boiling water	radiator fan, hair dryer, blood circulation

The rate of convective loss from a surface is roughly proportional to the excess temperature, H = hA(\theta-\theta_0), where h is the convection coefficient. This is precisely Newton's law of cooling from Unit 11 — Newton's law is a law of convection, which is why it fails when radiation takes over at high temperature.

Convection explains a long list of everyday things: the up-valley wind that blows through Pokhara and Kathmandu on a sunny afternoon and reverses at night; land and sea breezes; the draught up a chimney; the circulation of water in a solar water heater on a rooftop, which needs no pump because hot water rises on its own; why a room heater is placed low and an air conditioner high; and why the trapped air in a woollen sweater keeps you warm — the wool stops the air from convecting away.

12.3 Radiation: Ideal radiator

The Earth receives energy from the Sun across 150 million kilometres of vacuum, so there must be a third mechanism that needs no matter at all.

Radiation travels at 3\times10^{8}\ \text{m s}^{-1}, obeys the inverse-square law, and can be reflected and refracted like light. Prevost's theory of heat exchange states that every body above absolute zero radiates continuously, whatever its surroundings; a body appears to cool only because it emits more than it absorbs, and at thermal equilibrium the two rates are equal.

When radiation falls on a body, part is absorbed, part reflected and part transmitted. Writing each as a fraction of the incident energy,

That the best absorber is also the best emitter is Kirchhoff's law: at a given temperature and wavelength, the ratio of emissive power to absorptive power is the same for all bodies and equals the emissive power of a black body. The everyday proof is a glazed white cup with a coloured pattern: heat it in a furnace and the pattern, which absorbs best, glows brightest.

No real surface is perfectly black — lamp black absorbs about 96 %. A practical black body is made instead as a cavity: Ferry's black body is a double-walled copper sphere, blackened with lamp black inside, with a small hole and a conical projection opposite it. Radiation entering the hole is reflected many times inside and absorbed almost completely before it can find its way out, so the hole behaves as a black surface. Heat the sphere and the same hole becomes an ideal radiator.

The emissivity e of a real surface is the ratio of its emissive power to that of a black body at the same temperature, so 0 \le e \le 1, with e = 1 for a black body. By Kirchhoff's law e = a.

12.4 Black-body radiation

A black body does not radiate at one wavelength only. Measuring how the emitted power is distributed over wavelength gives the black-body spectrum, first measured accurately by Lummer and Pringsheim.

Four features of these curves must be quoted:

1. The spectrum is continuous: energy is emitted at every wavelength.
2. Each curve rises to a single maximum at a wavelength \lambda_m and falls away on both sides; the emission is very small at very short and very long wavelengths.
3. As the temperature rises, \lambda_m shifts towards shorter wavelengths. This is Wien's displacement law:

4. The area under a curve is the total power radiated per unit area, and it increases very rapidly with temperature — as T^4, which is the next section.

Wien's law is why a heated iron bar glows dull red, then orange, then white as it gets hotter, and why the blue star Rigel is hotter than the red star Betelgeuse. Classical physics could not reproduce these curves at all; it predicted that the emission should grow without limit at short wavelengths (the ultraviolet catastrophe). Planck fixed it in 1900 by assuming that energy is emitted in discrete quanta E = h\nu — the beginning of quantum theory.

12.5 Stefan-Boltzmann law

For a body of surface area A and emissivity e the total power radiated is P = e\sigma AT^{4}. But the body is also absorbing from its surroundings at T_0, at a rate e\sigma AT_0^{4} by Prevost's theory. The net rate of loss is therefore

A vacuum flask now makes complete sense. The double wall with the space evacuated stops conduction and convection; the silvered surfaces have a very low emissivity and so radiate and absorb very little; and the narrow insulating stopper blocks the only remaining conduction path.

Chapter summary

• Steady conduction: H = kA(\theta_1-\theta_2)/L = -kA\,d\theta/dx; thermal conductivity k is in W m⁻¹ K⁻¹.
• Thermal resistance R = L/(kA); resistances of slabs in series add, and H = \Delta\theta/R.
• Searle's method measures k for good conductors: k = mc_w(\theta_4-\theta_3)x/[A(\theta_1-\theta_2)t]; Lee's disc is used for bad conductors.
• Convection needs a moving fluid; natural convection is driven by density differences, forced convection by a fan or pump. Newton's law of cooling is a convection law.
• Radiation needs no medium. a+r+t=1; a perfect black body has a = e = 1; Kirchhoff's law says good absorbers are good emitters; Prevost's theory says emission never stops.
• Black-body curves are continuous with one maximum; Wien's displacement law gives \lambda_m T = 2.898\times10^{-3} m K.
• Stefan-Boltzmann law: E = \sigma T^4 with \sigma = 5.67\times10^{-8}\ \text{W m}^{-2}\text{K}^{-4}; net loss P = e\sigma A(T^4-T_0^4), with T in kelvin.

Practice questions

Group A — Multiple choice (1 mark each)

1. The SI unit of thermal conductivity is [1]aW m⁻¹ K⁻¹bW m⁻² K⁻⁴cJ kg⁻¹ K⁻¹dW m K
2. If the absolute temperature of a black body is doubled, the energy it radiates per second becomes [1]a2 timesb4 timesc8 timesd16 times
3. A good absorber of radiation is also a good emitter. This is [1]aStefan's lawbWien's lawcKirchhoff's lawdPrevost's theory
4. Heat transfer by convection cannot take place in [1]awaterbairca soliddany fluid
5. The wavelength of maximum emission from a black body at 2000 K is about 1450 nm. At 4000 K it will be about [1]a5800 nmb2900 nmc725 nmd1450 nm

Answers to Group A

1. (a) — from H = kA\Delta\theta/L, k has units W m⁻¹ K⁻¹.

2. (d) — E \propto T^4, and 2^4 = 16.

3. (c) — Kirchhoff's law, e = a.

4. (c) — convection needs the material itself to move, which a solid cannot do.

5. (c) — \lambda_m \propto 1/T, so doubling T halves \lambda_m.

Group B — Short answer (5 marks each)

1. Define thermal conductivity and derive H = kA(\theta_1-\theta_2)/L for steady conduction through a slab. Why must the slab be lagged? [5]
2. A wall of area 10\ \text{m}^2 is made of 12 cm of brick (k = 0.60\ \text{W m}^{-1}\text{K}^{-1}) covered on the outside with 4.0 cm of insulating board (k = 0.050\ \text{W m}^{-1}\text{K}^{-1}). The inside surface is at 25\ ^\circC and the outside surface at 5\ ^\circC. Find the rate of heat flow and the temperature of the junction. [5]
3. Distinguish between natural and forced convection, and explain why a thermos flask has a vacuum between its walls and silvered surfaces. [5]
4. A black body of surface area 20\ \text{cm}^2 is kept at 527\ ^\circC in surroundings at 27\ ^\circC. Find the net rate of loss of heat by radiation. [5]

Answers to Group B

1. Outline: define k as in §12.1; state H \propto A and H \propto \Delta\theta/L in the steady state; combine to H = kA\Delta\theta/L. Lagging ensures no heat escapes through the sides, so the heat current is the same at every cross-section and the gradient is uniform.

2. R_1 = L_1/(k_1A) = 0.12/(0.60\times10) = 0.020\ \text{K W}^{-1} and R_2 = 0.040/(0.050\times10) = 0.080\ \text{K W}^{-1}, so R = 0.100\ \text{K W}^{-1}. Then H = \Delta\theta/R = 20/0.100 = 200 W. Junction temperature \theta = 25 - HR_1 = 25 - 200\times0.020 = 21\ ^\circC. (Check: (21-5)/0.080 = 200 W.)

3. Natural convection is driven by the density change of the fluid itself in a gravitational field; forced convection is driven by an external pump or fan and is much faster. In a flask the vacuum removes the matter that conduction and convection need, and the silvering gives a very low emissivity so that little is radiated or absorbed.

4. A = 20\times10^{-4} = 2.0\times10^{-3}\ \text{m}^2, T = 800 K, T_0 = 300 K. T^4 - T_0^4 = 4.096\times10^{11} - 8.1\times10^{9} = 4.015\times10^{11}. So P = 5.67\times10^{-8}\times2.0\times10^{-3}\times4.015\times10^{11} = 45.5 W.

Group C — Long answer (8 marks each)

1. (a) Describe Searle's method for determining the thermal conductivity of a good conductor, with a labelled diagram, and derive the working formula. [6](b) State two reasons why the method is unsuitable for glass. [2]
2. (a) Draw the energy distribution curves of black-body radiation for three temperatures and state the conclusions drawn from them. [4](b) A star radiates most strongly at 350 nm and has a radius of 7.0\times10^{8} m. Find its surface temperature and its total power output, treating it as a black body. [4]

Answers to Group C

1. (a) Diagram and theory as in §12.1; the working formula is k = mc_w(\theta_4-\theta_3)x/[A(\theta_1-\theta_2)t], obtained by equating the heat conducted between the two bar thermometers to the heat carried away by the water. (b) Glass has a very small k, so the heat collected by the water in a reasonable time would be too small to measure, and the temperature difference \theta_4-\theta_3 would be lost in the experimental error; also, a long thin glass bar would lose most of the heat sideways despite lagging.

2. (a) Curves as in Figure 4, with the four conclusions of §12.4. (b) T = b/\lambda_m = 2.898\times10^{-3}/(350\times10^{-9}) = 8280 K. Surface area A = 4\pi R^2 = 4\pi(7.0\times10^{8})^2 = 6.16\times10^{18}\ \text{m}^2. With T^4 = (8280)^4 = 4.70\times10^{15}, P = \sigma AT^4 = 5.67\times10^{-8}\times6.16\times10^{18}\times4.70\times10^{15} = 1.6\times10^{27}\ \text{W}.