Heat transfer — conduction (Fourier's law k), convection, radiation (Stefan-Boltzmann σT⁴): Exercises

Definition One symbol for heat flow, one sign convention

Throughout this page we write every rate of heat transfer — conduction, convection, or radiation — as the same symbol $H$ (the heat current, units watts). The parent note happens to call the radiated rate $P$ ; here we rename it $H$ so that when we add channels (conduction + convection + radiation) we are always adding like-for-like watts.

Sign convention: $H > 0$ means heat flowing away from the hot body toward the cooler surroundings (hot → cold, as the Second law of thermodynamics demands). A negative $H$ would mean net heating of the body.
Area convention: $A$ always means the area through which that particular channel actually transfers heat. When a body radiates or convects from more than one face, we spell that out with a subscript, e.g. $A_{rad}$ = total radiating area (all faces summed), $A_{conv}$ = total convecting area.

Level 1 — Recognition

L1.1 — Pick the mode

For each situation name the dominant heat-transfer mode: (a) The Sun warming your face on a winter day. (b) A metal spoon getting hot in a mug of tea. (c) Warm water rising in a kettle before it boils.

Recall Solution

WHAT we check: what carries the energy in each case.

(a) Radiation — energy crosses the vacuum of space as electromagnetic waves; no medium exists between Sun and Earth.
(b) Conduction — the spoon is a solid, so atoms/electrons pass energy along without the metal itself flowing.
(c) Convection — the heated water becomes less dense, rises, and cold water sinks to replace it, a physically moving loop of fluid.

L1.2 — Read Fourier's law

State the SI units of thermal conductivity $k$ , and say whether raising the thickness $Δ x$ of a wall increases or decreases the heat current $H$ .

Recall Solution

From $H = k A \frac{Δ T}{Δ x}$ , solve for $k$ : $k = \frac{H Δ x}{A Δ T}$ . Units: $\frac{W \cdot m}{m ^{2} \cdot K} = W m^{- 1} K^{- 1}$ . Because $Δ x$ sits in the denominator, a thicker wall gives a smaller $H$ — a thicker wall insulates better. This matches the intuition "long path = slow flow."

Level 2 — Application

L2.1 — Heat through a brick wall

A brick wall has $k = 0.6 W m^{- 1} K^{- 1}$ , area $A = 15 m^{2}$ , thickness $Δ x = 0.20 m$ . Inside is $2 2^{\circ}$ C, outside is $2^{\circ}$ C. Find the heat current $H$ .

Figure — Heat transfer — conduction (Fourier's law k), convection, radiation (Stefan-Boltzmann σT⁴)

The figure above shows what "heat current" looks like: parallel arrows (a heat-flow field) streaming from the warm inside face to the cold outside face, and a colour gradient fading from magenta-hot to navy-cold. The arrows are all the same length because in steady state the same current $H$ crosses every plane of the wall — nothing piles up.

Recall Solution

WHAT: direct plug-in of Fourier's law for a single slab. WHY no kelvin conversion: we only need a temperature difference, and a difference in $^{\circ}$ C equals the same difference in K ( $Δ T = 22 - 2 = 20 K$ ). $H = k A \frac{Δ T}{Δ x} = 0.6 \times 15 \times \frac{20}{0.20} = 900 W$ The sign is positive, so by our convention heat flows out (inside → outside) — exactly as expected when it is warmer indoors.

L2.2 — Radiating sphere

A blackbody sphere ( $e = 1$ ) of radius $r = 0.05 m$ sits at $T = 400 K$ in surroundings at $T_{0} = 300 K$ . Find the net radiated heat current.

Recall Solution

WHAT: surface area of a sphere is $A = 4 π r^{2}$ , then apply net Stefan–Boltzmann (all radiating from the one closed surface). $A = 4 π (0.05)^{2} = 4 π \times 0.0025 = 0.031416 m^{2}$ WHY net, not just $T^{4}$ : the sphere both emits ( $e σ A T^{4}$ ) and absorbs from the room ( $e σ A T_{0}^{4}$ ); the observable heat loss is the difference. $H = 1 \times 5.67 \times 1 0^{- 8} \times 0.031416 \times (40 0^{4} - 30 0^{4})$ $40 0^{4} = 2.56 \times 1 0^{10}$ , $30 0^{4} = 8.1 \times 1 0^{9}$ , difference $= 1.75 \times 1 0^{10}$ . $H = 5.67 \times 1 0^{- 8} \times 0.031416 \times 1.75 \times 1 0^{10} \approx 31.2 W$ $H > 0$ : the hotter sphere loses heat to the cooler room, consistent with our sign rule.

Level 3 — Analysis

L3.1 — Composite (series) wall

A wall is two slabs of equal area $A = 10 m^{2}$ back-to-back: slab 1 (brick): $k_{1} = 0.6$ , $Δ x_{1} = 0.10 m$ ; slab 2 (foam): $k_{2} = 0.04$ , $Δ x_{2} = 0.05 m$ . Inside $2 0^{\circ}$ C, outside $0^{\circ}$ C. Find $H$ and the temperature at the brick–foam interface.

Read the figure like a story: the magenta line is the actual temperature as you walk from the hot inside face (left) to the cold outside face (right). It falls gently across the brick (low resistance = small drop) and then plunges across the foam (high resistance = big drop). The same-length flow arrows underneath remind you the current $H$ is identical in both slabs — the slope changes, not the flow.

Recall Solution

WHY resistances add: in steady state no energy piles up between the slabs, so the same heat current $H$ flows through both — exactly like resistors in series (see the Electrical resistance Ohm's law analogy). Series ⇒ add resistances. $R_{1} = \frac{Δ x _{1}}{k _{1} A} = \frac{0.10}{0.6 \times 10} = 0.016667 K/W$ $R_{2} = \frac{Δ x _{2}}{k _{2} A} = \frac{0.05}{0.04 \times 10} = 0.125 K/W$ $R_{tot} = 0.141667 K/W, H = \frac{Δ T}{R _{tot}} = \frac{20}{0.141667} \approx 141.2 W$ Interface temperature: the same $H$ crosses slab 1, so the temperature drop across brick is $Δ T_{1} = H R_{1} = 141.2 \times 0.016667 = 2.35 K$ . Starting from the hot face ( $2 0^{\circ}$ C): $T_{interface} = 20 - 2.35 \approx 17. 6^{\circ} C$ Notice the foam (huge $R_{2}$ ) carries almost the whole drop — that is what an insulator does: hog the temperature difference. This is exactly the plunge you see in the figure.

L3.2 — Which one wins the drop?

In L3.1, what fraction of the total $20 K$ drop happens across the foam? Interpret.

Recall Solution

$Δ T_{2} = H R_{2} = 141.2 \times 0.125 = 17.65 K$ . Fraction $= 17.65/20 = 0.88 = 88%$ . Interpretation: the slab with the larger resistance takes the larger temperature drop — just like the larger resistor drops more voltage in a series circuit. Insulation works by concentrating $Δ T$ inside itself.

Level 4 — Synthesis

L4.1 — Conduction feeding radiation (steady state)

A thin blackbody plate has geometric face area $A_{face} = 0.02 m^{2}$ and radiates from both faces, so its total radiating area is $A_{rad} = 2 A_{face} = 0.04 m^{2}$ . It is heated on its back by a conductive rod. In steady state the plate sits at $T = 500 K$ , surroundings at $T_{0} = 300 K$ , $e = 1$ . What conductive heat current $H$ must the rod deliver?

Recall Solution

WHY: in steady state energy in = energy out. The rod's conduction in must equal the net radiation out from both faces, else the plate's temperature would drift. $H = e σ A_{rad} (T^{4} - T_{0}^{4}), A_{rad} = 2 A_{face} = 0.04 m^{2}$ $50 0^{4} = 6.25 \times 1 0^{10}$ , $30 0^{4} = 8.1 \times 1 0^{9}$ , difference $= 5.44 \times 1 0^{10}$ . $H = 1 \times 5.67 \times 1 0^{- 8} \times 0.04 \times 5.44 \times 1 0^{10} \approx 123.4 W$ Sanity: if the rod delivered more, the plate would heat up until its $T^{4}$ radiation rose to match; if less, it would cool. Steady state is the balance point.

L4.2 — Combined convection + radiation loss

A freestanding vertical panel at $T_{s} = 350 K$ has geometric face area $A_{face} = 1 m^{2}$ per face and emissivity $e = 0.9$ . Air and surroundings are both at $T_{0} = 290 K$ , and the convection coefficient is $h = 8 W m^{- 2} K^{- 1}$ . Because the panel stands free in the air, both faces radiate and convect. Find the total heat lost.

Recall Solution

WHY both faces: a freestanding panel is exposed to the room on the front and the back, so the effective areas double: $A_{rad} = A_{conv} = 2 A_{face} = 2 m^{2} .$ WHY add the two channels: convection and radiation act in parallel — two independent channels draining heat from the same surface — so their currents add. Convection: $H_{c} = h A_{conv} (T_{s} - T_{0}) = 8 \times 2 \times (350 - 290) = 960 W$ . Radiation: $35 0^{4} = 1.50063 \times 1 0^{10}$ , $29 0^{4} = 7.0728 \times 1 0^{9}$ , difference $= 7.9335 \times 1 0^{9}$ . $H_{r} = 0.9 \times 5.67 \times 1 0^{- 8} \times 2 \times 7.9335 \times 1 0^{9} = 809.6 W$ $H_{total} = H_{c} + H_{r} = 960 + 809.6 \approx 1769.6 W$ Both channels matter here — neither is negligible, so you cannot use the "drop the small term" shortcut. All terms are positive: the hot panel loses heat to the cooler room on both sides.

Level 5 — Mastery

L5.1 — Temperature-drop bookkeeping across a window with a still-air film

A double-glazed unit is: glass ( $k_{g} = 0.8$ , $Δ x_{g} = 0.004 m$ ) + air gap ( $k_{a} = 0.025$ , $Δ x_{a} = 0.012 m$ ) + glass ( $k_{g} = 0.8$ , $Δ x_{g} = 0.004 m$ ), all area $A = 2 m^{2}$ . Inside $2 0^{\circ}$ C, outside $0^{\circ}$ C. Find $H$ and confirm the air gap does the insulating.

The figure draws the three resistances as blocks whose width is proportional to their resistance: the two glass panes are hair-thin, the trapped-air block is enormous. Visually, the air gap is nearly the whole "electrical" length of the stack — so it must eat nearly the whole temperature drop.

Recall Solution

WHAT: three slabs in series ⇒ three resistances added. $R_{g} = \frac{0.004}{0.8 \times 2} = 0.0025 K/W (each glass pane)$ $R_{a} = \frac{0.012}{0.025 \times 2} = 0.24 K/W (air gap)$ $R_{tot} = 0.0025 + 0.24 + 0.0025 = 0.245 K/W$ $H = \frac{20}{0.245} \approx 81.6 W$ Compare with the single 4 mm pane from the parent note ( $8000 W$ ): the air gap cut the loss ~98%. Why the air gap dominates: its $R_{a} = 0.24$ is ~96× each glass $R_{g} = 0.0025$ . Temperature drop across the air gap: $Δ T_{a} = H R_{a} = 81.6 \times 0.24 = 19.6 K$ — almost the whole $20 K$ sits in the trapped air (which is why the air must be still, so convection can't short-circuit it).

L5.2 — Design: match a radiator to a room's cooling

A hot object ( $e = 1$ , radiating area $A_{rad} = 0.05 m^{2}$ ) must radiate a net $200 W$ to surroundings at $T_{0} = 300 K$ . What surface temperature $T$ is required?

Recall Solution

WHY invert: we know the target heat current and solve back for $T$ — the 4th-power law inverts with a fourth root. $H = e σ A_{rad} (T^{4} - T_{0}^{4}) \Rightarrow T^{4} = T_{0}^{4} + \frac{H}{e σ A _{rad}}$ $\frac{H}{e σ A _{rad}} = \frac{200}{1 \times 5.67 \times 1 0 ^{- 8} \times 0.05} = 7.0547 \times 1 0^{10}$ $T^{4} = 8.1 \times 1 0^{9} + 7.0547 \times 1 0^{10} = 7.8647 \times 1 0^{10}$ $T = (7.8647 \times 1 0^{10})^{1/4} \approx 529.4 K$ Check the physics: $T > T_{0}$ as it must be for net outward radiation ( $H > 0$ , hot → cold; see Second law of thermodynamics).

Recall Quick self-check ladder

One-line reason ::: L1 mode = mechanism; L2 kelvin only for powers; L3 series adds resistances; L4 both faces + channels add in parallel; L5 invert the net law with $+ T_{0}^{4}$ inside the root.

Related deep threads: Newton's law of cooling, Black body radiation, Wien's displacement law, Planck's law, Greenhouse effect, Thermal conductivity k.