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Fourier's law of conduction, thermal resistances in series, convection, and the Stefan-Boltzmann law for radiation, applied to walls, windows, insulation and people.
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By the end of this lesson you will be able to find a conduction rate with Fourier's law, combine the thermal resistances of layers in series and find each layer's temperature drop, convert and use insulation R-values, and find the net radiation between a surface and its surroundings.
You know that heat flows from hot to cold until the temperatures are equal, and you have met conduction, convection and radiation as mechanisms. You also know Ohm's law and that resistors in series add. This lesson makes the mechanisms quantitative, and discovers that a wall conducting heat obeys the same algebra as a circuit conducting charge.
| Term | What it means |
|---|---|
| Thermal conductivity | $k$, in W/(m K): how well a material conducts heat. Copper $400$, glass $0.8$, fiberglass $0.04$, still air $0.026$. |
| Fourier's law | $\dot{Q} = kA\,\Delta T/L$: conduction rate through a slab of area $A$ and thickness $L$. |
| Heat flux | Heat per unit time per unit area, $q = \dot{Q}/A$, in W/m². |
| Thermal resistance | $R = L/(kA)$ in K/W for a slab; per unit area $L/k$ in m² K/W. Layers in series add. |
| R-value | The resistance per unit area of an insulation product; US values are in ft² °F h/BTU, and one US unit is $0.1761$ m² K/W. |
| Convection coefficient | $h$, in W/(m² K): the rate of heat carried by a moving fluid per unit area and per kelvin of difference. |
| Emissivity | $\varepsilon$, between $0$ and $1$: how nearly a surface radiates like a perfect blackbody. |
| Stefan-Boltzmann constant | $\sigma = 5.67 \times 10^{-8}$ W/(m² K⁴). |
Conduction. Put a slab of material between a warm face and a cool one. Energy passes from faster molecules to slower ones, layer by layer, and in the steady state the rate is
$$\dot{Q} = kA\frac{\Delta T}{L},$$
Fourier's law. The rate is proportional to the area $A$, to the temperature difference $\Delta T$, and inversely to the thickness $L$; the thermal conductivity $k$ carries the material. Metals conduct well because their free electrons carry energy; gases conduct badly because their molecules are far apart, which is why nearly every insulator — fiberglass, foam, down, wool — works by trapping still air.
Thermal resistance. Rewrite Fourier's law as $\dot{Q} = \Delta T/R$ with $R = L/(kA)$. This is Ohm's law with heat flow for current and temperature difference for voltage. Layers in series carry the same heat flow, so their resistances add, and each takes a share of the temperature drop in proportion to its resistance.
Convection. A fluid moving past a surface carries energy away, at a rate $\dot{Q} = hA\,\Delta T$ where the coefficient $h$ depends on the flow: about $10$ W/(m² K) for still indoor air, $30$ or more in a wind.
Radiation. Every surface emits electromagnetic radiation at a rate
$$P = \varepsilon\sigma AT^4,$$
the Stefan-Boltzmann law, with $T$ absolute. A surface also absorbs radiation from its surroundings, so the net exchange with surroundings at $T_0$ is $\varepsilon\sigma A(T^4 - T_0^4)$. The fourth power makes radiation small near room temperature and overwhelming at high temperature: doubling the absolute temperature multiplies the emission by sixteen.
Another way: picture
Picture heat moving through a wall like water through a series of narrow pipes. The temperature difference is the pressure pushing it; each layer is a pipe with its own resistance; the flow is the same through every pipe, and the pressure drops most across the narrowest one. In a well-insulated wall the fiberglass is the narrow pipe, and the brick and drywall hardly matter.
Another way: steps
Every problem here reduces to a rate equals a driving difference over a resistance, and the method is to build the resistance correctly.
Checks. The individual temperature drops must add up to the whole difference. The largest resistance takes the largest share. A radiation answer must use kelvin — putting in $200$ °C instead of $473$ K underestimates the emission by a factor of thirty. And a conduction answer in thousands of watts through a single window is a sign that the air films have been left out: they, not the glass, provide most of a single pane's resistance.
A single pane of glass $4$ mm thick has a resistance per square meter of $0.004/0.80 = 0.005$ m² K/W. With a $20$ K difference across the glass itself it would pass $4000$ W per square meter — the output of two space heaters through one window. Real single-pane windows lose far less, because the glass is not what holds the temperature difference.
On each side of the pane sits a thin layer of air that the glass drags almost to a standstill. Still air conducts poorly, and these films add resistances of roughly $0.12$ m² K/W inside and $0.03$ m² K/W outside, where the wind thins them. The total is $0.12 + 0.005 + 0.03 = 0.155$ m² K/W, and the flux for a $20$ K indoor-outdoor difference is about $130$ W/m². Almost the whole temperature drop is across the air; the glass surface is nearly at the outdoor temperature, which is why it fogs and frosts.
A double-glazed window adds a sealed layer of gas between two panes, about $0.17$ m² K/W more, which roughly halves the loss. Filling the gap with argon, whose conductivity is two thirds of air's, and coating one pane to reflect infrared radiation cuts it further. Every one of those improvements is a resistance added in series.
The Stefan-Boltzmann law comes from the physics of blackbody radiation that closes this course: the energy of the photon gas in a cavity grows as $T^4$, and a surface emits in proportion to it. For now, the consequences.
A person with $1.8$ m² of skin at $306$ K, emissivity $0.97$, emits $0.97 \times 5.67 \times 10^{-8} \times 1.8 \times 306^4 \approx 868$ W — more than eight times the body's resting heat output. It survives because the walls of a $293$ K room send back about $730$ W, and the net loss is about $140$ W. That net loss depends on the walls, not the air: sitting near a cold window on a winter night feels chilly even in a warm room, because one side of the body is exchanging radiation with a surface many degrees colder.
At high temperature the fourth power takes over. A stove surface at $600$ K emits sixteen times as much per square meter as the same surface at $300$ K; the Sun's surface at $5800$ K emits about $64$ MW from every square meter.
Rolls of fiberglass in American hardware stores are labeled R-13, R-19, R-30, R-38 and R-49. The number is the thermal resistance of one square foot, in ft² °F h/BTU, and the Department of Energy recommends R-49 to R-60 for attics in the coldest northern states.
Convert R-38 to SI: $38 \times 0.1761 = 6.69$ m² K/W. A $100$ m² ceiling insulated to it has a conductance of $100/6.69 = 14.9$ W/K. On a January night in Minneapolis, with the rooms at $20$ °C and the attic at $-20$ °C, the difference is $40$ K and the loss through the ceiling is $14.9 \times 40 \approx 600$ W: about $52$ MJ, or $14$ kWh, a day.
The same ceiling at R-13, typical of houses built before the 1970s, has resistance $2.29$ m² K/W and loses $100 \times 40/2.29 \approx 1750$ W, nearly three times as much. Because the flow goes as $1/R$, each extra unit of R saves less than the last: going from R-13 to R-38 saves about $1150$ W, while going from R-38 to R-60 saves only about $220$ W more. That diminishing return is why the recommended value depends on the climate — insulation pays for itself fastest where the temperature difference is largest.
A vacuum flask keeps coffee hot for a whole day by attacking all three mechanisms at once. The coffee sits in a glass or steel vessel inside a second one, with the gap between them pumped out. With no gas in the gap there is nothing to conduct or convect heat across it; the only conduction path left is the thin neck where the two walls join, which is made long and narrow to give it a large resistance.
That leaves radiation, which crosses a vacuum perfectly well. The walls facing the gap are silvered, dropping their emissivity to about $0.02$. Radiation between two surfaces at $360$ K and $295$ K, each $0.05$ m² with that emissivity, carries roughly $0.01 \times 5.67 \times 10^{-8} \times 0.05 \times (360^4 - 295^4) \approx 0.3$ W, since two silvered surfaces facing each other behave like one of emissivity about $0.01$. Half a liter of coffee holds about $2100$ J per kelvin, so at that rate it would lose only about half a kelvin an hour through the walls. In practice the neck and stopper lose more, and a good flask drops from $90$ °C to about $70$ °C in a day — still a remarkable result for a device with no moving parts and no power supply.
When layers are stacked, it is tempting to think that each contributes equally or that adding any layer helps a lot. In series the resistances add, so the layer with by far the largest resistance controls the flow, and a highly conducting layer — a sheet of metal, a brick facing in front of good insulation — adds almost nothing. Doubling the insulation's thickness nearly halves the loss; doubling the brick barely changes it.
The second error is to picture radiation as something only a hot object does, and only toward colder things. Every surface above absolute zero radiates, all the time, at a rate set by its own temperature. Two objects at the same temperature exchange radiation continuously; they simply send each other equal amounts, so the net flow is zero. That is why the net radiation formula has a difference of fourth powers, and why a cold wall can chill you across a warm room.
A pane of glass $1.5$ m² in area and $4.0$ mm thick has surfaces $20$ K apart. Write Fourier's law.
$\dot{Q} = \dfrac{kA\,\Delta T}{L}$
Steady conduction through a slab.
Put the thickness in meters.
$L = 4.0\ \text{mm} = 0.0040\ \text{m}$
The conductivity is in watts per meter kelvin.
Substitute, with $k = 0.80$ W/(m K) for glass.
$\dot{Q} = \dfrac{0.80 \times 1.5 \times 20}{0.0040}$
Area in square meters and temperature difference in kelvin.
Work out the rate.
$\dot{Q} = \dfrac{24}{0.0040} = 6000\ \text{W}$
A surprisingly large number.
Judge it against the air films.
$R_{\text{glass}} = \dfrac{0.0040}{0.80} = 0.005 \ll R_{\text{films}} \approx 0.15\ \text{m}^2\,\text{K/W}$
Twenty kelvin across the glass alone never happens: the air films on either side take almost all of the temperature difference.
A wall is $0.10$ m of brick ($k = 0.70$) and $0.089$ m of fiberglass ($k = 0.040$). Find the brick's resistance per square meter.
$R_1 = \dfrac{0.10}{0.70} \approx 0.143\ \text{m}^2\,\text{K/W}$
Thickness over conductivity, for one square meter.
Find the fiberglass's resistance per square meter.
$R_2 = \dfrac{0.089}{0.040} \approx 2.225\ \text{m}^2\,\text{K/W}$
Three and a half inches of fiberglass, the standard fill of a stud wall.
Add the two resistances.
$R = 0.143 + 2.225 = 2.368\ \text{m}^2\,\text{K/W}$
Series layers add their resistances.
Find the flux for a $25$ K difference.
$q = \dfrac{\Delta T}{R} = \dfrac{25}{2.368} \approx 10.6\ \text{W/m}^2$
The same flux crosses both layers.
Find the loss through $40$ m² of wall.
$\dot{Q} = qA = 10.6 \times 40 \approx 422\ \text{W}$
About a quarter of a small space heater's output.
Find the temperature drop across the brick.
$\Delta T_1 = qR_1 = 10.6 \times 0.143 \approx 1.5\ \text{K}$
The brick takes only six percent of the drop; the fiberglass takes the rest, because it holds nearly all the resistance.
Skin of area $1.8$ m² and emissivity $0.97$ is at $33$ °C; the room's walls are at $20$ °C. Convert both temperatures to kelvin.
$T = 33 + 273 = 306\ \text{K}, \qquad T_0 = 20 + 273 = 293\ \text{K}$
The Stefan-Boltzmann law needs absolute temperatures.
Work out the common factor.
$\varepsilon\sigma A = 0.97 \times 5.67 \times 10^{-8} \times 1.8 = 9.90 \times 10^{-8}\ \text{W/K}^4$
It multiplies both the emission and the absorption.
Find the power the skin emits.
$P_{\text{emit}} = 9.90 \times 10^{-8} \times 306^4 = 9.90 \times 10^{-8} \times 8.77 \times 10^{9} \approx 868\ \text{W}$
Every surface radiates at its own temperature.
Find the power it absorbs from the walls.
$P_{\text{abs}} = 9.90 \times 10^{-8} \times 293^4 = 9.90 \times 10^{-8} \times 7.37 \times 10^{9} \approx 730\ \text{W}$
A good emitter is an equally good absorber, so the same factor applies.
Subtract for the net loss.
$P_{\text{net}} = 868 - 730 \approx 138\ \text{W}$
The net exchange is what the body must make up.
Compare with the body's heat production.
$P_{\text{resting}} \approx 100\ \text{W} < 138\ \text{W}$
Unclothed, the body would lose heat faster than it makes it, which is why clothing, a second resistance, is needed indoors.
Linearize for a small difference as a check.
$P_{\text{net}} \approx 4\varepsilon\sigma AT_0^3\,\Delta T = 4 \times 9.90 \times 10^{-8} \times 293^3 \times 13 \approx 130\ \text{W}$
For small differences $T^4 - T_0^4 \approx 4T_0^3\Delta T$, so radiation behaves like conduction with an effective coefficient.
Find the resistance.
$R = \dfrac{L}{kA} = \dfrac{0.15}{0.040 \times 1.0} = 3.75\ \text{K/W}$
Thickness over conductivity times area.
Divide the temperature difference by the resistance.
$\dot{Q} = \dfrac{\Delta T}{R} = \dfrac{30}{3.75}$
Heat flow is driving difference over resistance.
Work out the rate.
A slab of insulation conducts heat at some rate. It is replaced by the same material with $2$ times the area and $3$ times the thickness, with the same temperatures on its faces. By what factor does the rate of conduction change?
Complete the worked solution: a wall is brick with a resistance of $0.25$ m² K/W per square meter, plus $7$ cm of foam board ($k = 0.040$ W/(m K)). The two faces differ by $18$ K. What heat flux crosses it?
Find the foam's resistance per square meter.
$R_2 = \dfrac{L}{k} = \dfrac{0.07}{0.040} =$ a m² K/W
The thickness is in meters, $7$ cm $= 0.07$ m.
Add the two resistances.
$R = 0.25 + R_2 =$ b m² K/W
The same heat crosses both layers in turn.
Divide the temperature difference by the total resistance.
$q = \dfrac{18}{R} =$ f W/m²
Flux is the driving difference over the resistance, as current is voltage over resistance.
Match each way heat moves to the law that gives its rate, as used in a house with $3$ outside walls.
| $\dot{Q} = kA\,\Delta T/L$ | $\dot{Q} = hA\,\Delta T$ | $P = \varepsilon\sigma AT^4$ | resistances add: $R = R_1 + R_2 + R_3$ | |
|---|---|---|---|---|
| conduction through a solid | ||||
| convection into moving air | ||||
| radiation from a hot surface | ||||
| layers of a wall, one after another |
A wall is $0.10$ m of brick ($k = 0.50$ W/(m K)), then $0.090$ m of fiberglass ($k = 0.040$ W/(m K)), then $0.0125$ m of drywall ($k = 0.25$ W/(m K)). The inside surface is $34$ K warmer than the outside. Fill in each layer's resistance per square meter and its temperature drop.
| resistance (m² K/W) | temperature drop (K) | |
|---|---|---|
| brick | ||
| fiberglass | ||
| drywall | ||
| whole wall |
A pane of glass ($k = 0.80$ W/(m K)) has an area of $1.8$ m² and a thickness of $2$ mm. Its two surfaces differ in temperature by $1.3$ K. At what rate does heat conduct through it?
Answer: unit: W / kW / mW
The iron surface of a woodstove has an area of $1.2$ m² and an emissivity of $0.90$. It is at $500$ K in a room whose walls are at $300$ K. What net power does it radiate into the room? Use $\sigma = 5.67 \times 10^{-8}$ W/(m² K⁴).
Answer: unit: W / kW / mW
A house in Minnesota has $100$ m² of ceiling insulated to R-$49$ in US units. The rooms are $15$ K warmer than the attic. Ignoring the thin drywall, at what rate does heat escape through the ceiling?
Answer: unit: W / kW / mW
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A wall is $0.10$ m of brick ($k = 0.50$ W/(m K)), then $0.090$ m of fiberglass ($k = 0.040$ W/(m K)), then $0.0125$ m of drywall ($k = 0.25$ W/(m K)). The inside surface is $10$ K warmer than the outside. Fill in each layer's resistance per square meter and its temperature drop.
| resistance (m² K/W) | temperature drop (K) | |
|---|---|---|
| brick | ||
| fiberglass | ||
| drywall | ||
| whole wall |
You can find how fast heat crosses a layered wall and how much a hot surface radiates. Explain to someone why adding a brick facing to a well-insulated wall hardly changes its heat loss.
14. Your turn: a $1.0$ m² batt of fiberglass $0.15$ m thick ($k = 0.040$ W/(m K)) has a $30$ K difference across it. What is the rate of heat flow?, step 3
$\dot{Q} = 8.0\ \text{W}$
Thick, poorly conducting insulation lets through only a trickle.