Heat conduction: practice problems with step-by-step solutions
Free Heat conduction practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: single wall, composite wall, wall with convection.
Free practice problemEasy
A wall of expanded polystyrene is 10 mm thick and has an area of 6 m². Its faces are held at 24 °C and -11 °C. Find the rate of heat conduction Q̇ through the wall.
k = 0.035 W/(m·K)L = 10 mmA = 6 m²T1 = 24 °CT2 = -11 °C
Heat always flows from hot to cold. Inside a solid it travels by conduction: fast-vibrating particles on the warm side pass their energy on to their neighbours.
On a winter day, heat leaves a warm room through the walls in this way. How much leaks out depends on the temperature difference, the wall area, its thickness and the material.
Materials differ hugely. The thermal conductivityk, in W/(m·K), tells how easily heat passes through:
Concrete: about 1.4 W/(m·K). Brick: about 0.72.
Wood: about 0.17.
Mineral wool: about 0.04. Expanded polystyrene: about 0.035.
A low k means a good insulator. That is why houses have a layer of wool or foam inside the wall.
In this topic the temperatures do not change with time. This is called steady conduction.
Self-check Which wall lets the least heat through, if all are 100 mm thick?
Brick (k = 0.72)
Concrete (k = 1.4)
Mineral wool (k = 0.04)
Wood (k = 0.17)
Answer: Mineral wool (k = 0.04). The heat flow is proportional to k. Mineral wool has by far the lowest conductivity, so it insulates best.
2Fourier’s law and thermal resistance
For a flat wall of thickness L and area A, with its faces at T1 and T2, the rate of heat flow in watts is given by Fourier’s law:
Q̇ = k · A · (T1 − T2) / L
The temperature falls in a straight line through the wall. The heat flux is the flow per square metre: q″ = Q̇ / A, in W/m².
It helps to think of the wall as a resistor, just like in an electric circuit. The temperature difference drives the heat, and the wall resists it:
Q̇ = ΔT / R R = L / (k · A)
The thermal resistance R is in K/W. A thicker wall or a lower k gives a higher resistance and less heat loss. A temperature difference in °C is the same number in K.
Self-check A concrete wall (k = 1.4 W/(m·K)) is 200 mm thick with an area of 10 m². Its faces are at 20 °C and 0 °C. What is the heat flow?
0.0007 W
1.4 W
140 W
1400 W
Answer: 1400 W. Q̇ = 1.4 · 10 · 20 / 0.2 = 1400 W. Using L = 200 (mm) gives 1.4 W, and forgetting the area gives 140 W.
3Walls with several layers
A real wall has layers: brick, insulation, plasterboard. The same heat passes through each layer one after the other, so the layers are resistances in series. Add them:
Rtotal = R1 + R2 + … Q̇ = (Tin − Tout) / Rtotal
Each layer takes a share of the total temperature drop in proportion to its resistance: ΔT1 = Q̇ · R1. The insulation layer takes the biggest drop.
Air next to a wall also resists heat. This convection resistance is R = 1/(h · A), where h is the convection coefficient in W/(m²·K). Add it in series like a layer.
The U-value sums up a whole wall, including the air films, as heat flow per m² per kelvin:
U = 1 / (Rtotal · A) Q̇ = U · A · ΔT
4The method step by step
Draw the wall with its layers. Write each thickness in metres and each k next to it.
Calculate the resistance of every layer: R = L/(k·A). With air on a side, add R = 1/(h·A) for that side.
Add all the resistances in series.
Find the heat flow: Q̇ = (total temperature difference) / Rtotal.
For a temperature inside the wall, start from a known temperature and subtract Q̇ · R for each layer you cross.
A good check: the temperature drops across all the layers must add up to the total temperature difference.
5A worked example
Follow every step. Try to predict the next one before you open it.
Worked example
A wall is made of 200 mm of brick (k = 0.72 W/(m·K)) and 50 mm of mineral wool (k = 0.04 W/(m·K)), with an area of 10 m². The outer faces are at 20 °C and 0 °C. Find the heat flow through the wall and the temperature at the interface between the layers.
1
Draw the wall
Heat flows from the 20 °C side through the brick, then through the wool, to the 0 °C side.
2
Resistance of each layer
R1 = 0.200 / (0.72 · 10) = 0.0278 K/W
R2 = 0.050 / (0.04 · 10) = 0.125 K/W
The thin wool layer resists much more than the thick brick.
3
Total resistance and heat flow
Rtotal = 0.0278 + 0.125 = 0.1528 K/W
Q̇ = (20 − 0) / 0.1528 = 130.9 W
4
Interface temperature
Start at the warm face and cross the brick:
Ti = 20 − Q̇ · R1 = 20 − 130.9 · 0.0278 = 16.4 °C
5
Check the answer
The drop across the wool is 130.9 · 0.125 = 16.4 K, from 16.4 °C down to 0 °C. The two drops, 3.6 K + 16.4 K, add up to the full 20 K.
Answer
Q̇ ≈ 131 W, and the interface is at Ti ≈ 16.4 °C.
6Common mistakes
Leaving the thickness in mm. The k-values are per metre, so L must be in metres.
Averaging the k-values of two layers, or adding them. Add the resistances L/(k·A) instead.
Forgetting the area: R = L/(k·A) for the whole wall, R = L/k for one square metre.
Starting from the wrong side when you find an interface temperature. The temperature always drops in the direction of the heat flow.
Leaving out the air films (1/(h·A)) when inside and outside air temperatures are given.
Self-check One square metre of wall has 100 mm of mineral wool (k = 0.04) and 100 mm of brick (k = 0.72). The faces are 20 K apart. What is the heat flow?
152 W
38 W
7.6 W
8 W
Answer: 7.6 W. R = 0.1/0.04 + 0.1/0.72 = 2.5 + 0.139 = 2.64 K/W, so Q̇ = 20/2.64 = 7.6 W. Averaging k gives 38 W, and ignoring the brick gives 8 W.
7How exam questions look
Easy: a single wall. Find the heat flow Q̇ or the heat flux q″ with Fourier’s law.
Medium: a composite wall of two layers. Find the heat flow, or the temperature Ti at the interface between the layers.
Hard: a wall with convection on both sides, between room air and outside air. Find the U-value, the heat loss, or the temperature Ts of the inside surface.
For the inside surface temperature, start at the room air and cross only the inside air film: Ts = Troom − Q̇ / (hi · A).
Example: the wall above with h = 8 W/(m²·K) inside and 25 W/(m²·K) outside has Rtotal = 0.0125 + 0.1528 + 0.004 = 0.1693 K/W, so U = 1/(0.1693 · 10) = 0.59 W/(m²·K).
A wall of wood is 150 mm thick and has an area of 6 m². Its faces are held at 20 °C and -9 °C. Find the rate of heat conduction Q̇ through the wall.
k = 0.17 W/(m·K)L = 150 mmA = 6 m²T1 = 20 °CT2 = -9 °C
Find Q̇(W)
Show hints
Fourier’s law for a plane wall: Q̇ = k·A·ΔT/L.
Convert the thickness to metres so the units of k match.
ΔT = 29 K.
Show the answer and the step-by-step solution
Thermal resistance: steady heat flow through layers in series behaves like current through resistors in series: Q̇ = ΔT/ΣR, with R = L/(kA) for a conducting layer and R = 1/(hA) for a convecting surface. The temperature drop across each layer is proportional to its resistance.
Fourier’s law
ΔT = 20 − (-9) = 29 K
Q̇ = kAΔT/L = 0.17·6·29/(150/1000) = 197.2 W
Answer
Q̇ = 197.2 W
Example 2Medium
A wall is made of 50 mm of mineral wool and 50 mm of wood, with an area of 12 m². The outer faces are at 25 °C and 5 °C. Find the temperature Ti at the interface between the two layers.
L1 = 50 mm, k1 = 0.04 W/(m·K)L2 = 50 mm, k2 = 0.17 W/(m·K)A = 12 m²T1 = 25 °CT3 = 5 °C
Find Ti(°C)
Show hints
Each layer is a thermal resistance R = L/(kA). In series they add.
The same Q̇ flows through both layers, so Ti = T₁ − Q̇·R₁.
R1 + R2 = 0.1287 K/W.
Show the answer and the step-by-step solution
Thermal resistance: steady heat flow through layers in series behaves like current through resistors in series: Q̇ = ΔT/ΣR, with R = L/(kA) for a conducting layer and R = 1/(hA) for a convecting surface. The temperature drop across each layer is proportional to its resistance.
Layer resistances
R1 = L1/(k1A) = (50/1000)/(0.04·12) = 0.1042 K/W
R2 = L2/(k2A) = (50/1000)/(0.17·12) = 0.02451 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 20/0.1287 = 155.4 W
Interface temperature
The temperature drops across the first layer by Q̇·R₁.
Ti = T1 − Q̇·R1 = 25 − 155.4·0.1042 = 8.81 °C
Answer
Ti = 8.81 °C
Example 3Hard
A wall of 100 mm mineral wool and 25 mm gypsum board separates a room at 20 °C from outside air at -7 °C. The convection coefficients are 6 W/(m²·K) inside and 20 W/(m²·K) outside, and the wall area is 8 m². Find the overall heat transfer coefficient U of the wall.
L1 = 100 mm, k1 = 0.04 W/(m·K)L2 = 25 mm, k2 = 0.17 W/(m·K)A = 8 m²hi = 6 W/(m²·K)ho = 20 W/(m²·K)Troom = 20 °CTout = -7 °C
Find U(W/(m²·K))
Show hints
Add four resistances in series: inside convection, two conducting layers, outside convection.
U is the heat flow per unit area per degree: U = 1/(A·ΣR).
ΣR = 0.358 K/W.
Show the answer and the step-by-step solution
Thermal resistance: steady heat flow through layers in series behaves like current through resistors in series: Q̇ = ΔT/ΣR, with R = L/(kA) for a conducting layer and R = 1/(hA) for a convecting surface. The temperature drop across each layer is proportional to its resistance.
Resistances in series
Ri = 1/(hiA) = 1/(6·8) = 0.02083 K/W
R1 = L1/(k1A) = (100/1000)/(0.04·8) = 0.3125 K/W
R2 = L2/(k2A) = (25/1000)/(0.17·8) = 0.01838 K/W
Ro = 1/(hoA) = 1/(20·8) = 0.00625 K/W
ΣR = 0.358 K/W
Overall coefficient
U = 1/(A·ΣR) = 1/(8·0.358) = 0.3492 W/(m²·K)
Answer
U = 0.3492 W/(m²·K)
Questions and answers
What is Heat conduction?
Heat always flows from hot to cold. Inside a solid it travels by conduction: fast-vibrating particles on the warm side pass their energy on to their neighbours.
On a winter day, heat leaves a warm room through the walls in this way. How much leaks out depends on the temperature difference, the wall area, its thickness and the material.
How do I solve Heat conduction problems step by step?
Thermal resistance: steady heat flow through layers in series behaves like current through resistors in series: Q̇ = ΔT/ΣR, with R = L/(kA) for a conducting layer and R = 1/(hA) for a convecting surface. The temperature drop across each layer is proportional to its resistance.
What do the Heat conduction exercises cover?
Easy: single wall
Medium: composite wall
Hard: wall with convection
Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.
How practising works
Every problem has new values, so you learn the method instead of an answer.
Stuck? Open a small hint first, and the full explanation only when you need it.
Your progress, mistakes and saved problems are kept, so you can review before the exam.