The equation of rate of flow of heat describes how quickly heat energy moves through a material. It connects thermal conductivity, area, thickness, and temperature difference into one powerful formula that shows up in many physics questions and real-world designs.
Key takeaways
- The rate of flow of heat by conduction is often written as Q̇ = −k A (dT/dx) (Fourier’s law).
- For a uniform slab, the steady-state rate is Q/t = k A (Thot − Tcold) / L.
- k is thermal conductivity (W/m·K), A is area (m²), L is thickness (m), ΔT is temperature difference (K).
- Metals (like copper) have high k, so they conduct heat quickly; materials like wood and foam have low k, so they act as insulators.
- These equations are widely used in problems involving walls, rods, cooking pans, refrigerators, and building insulation.
Table of Contents
1. Core concept: rate of flow of heat
When one end of a solid is hotter than the other, heat energy flows from the hot region to the cold region. The rate of flow of heat tells us how many joules of heat energy pass through a surface each second.
In symbols, the rate of heat flow is often written as Q̇ (read “Q dot”) or Q/t, and it has the unit watt (W), which is the same as joule per second (J/s).
2. Standard equations & units
The fundamental law for conduction in a solid is Fourier’s law of heat conduction.
Q̇ = −k A &dfrac{dT}{dx}
Where:
• Q̇ = rate of heat flow (W or J/s)
• k = thermal conductivity (W/m·K)
• A = cross-sectional area perpendicular to heat flow (m²)
• dT/dx = temperature gradient along the direction of heat flow (K/m)
• The minus sign (−) indicates heat flows from higher temperature to lower temperature.
For a uniform slab of material of thickness L, with one side at temperature Thot and the other at Tcold, the steady-state rate of heat flow is:
&dfrac{Q}{t} = k A &dfrac{Thot − Tcold}{L}
• Q/t = rate of heat flow (W)
• Thot − Tcold = temperature difference across the slab (K or °C)
• L = thickness of the slab in the direction of heat flow (m)
Sometimes it is convenient to define a thermal resistance:
Rth = &dfrac{L}{k A} so that &dfrac{Q}{t} = &dfrac{Thot − Tcold}{Rth}
• Rth has units K/W (kelvin per watt).
• A large Rth means poor conduction (good insulation); a small Rth means good conduction.
If the rate of heat flow is roughly constant for a time interval t, then the total heat transferred is:
Q = &left(&dfrac{Q}{t}&right) t
This connects power (rate) and total energy in joules, and is often used in multi-step problems.
3. Thermal conductivity comparison table
Different materials conduct heat at very different rates. The table below shows approximate thermal conductivity values near room temperature.
| Material | Typical use | k (W/m·K) | Comment |
|---|---|---|---|
| Air | Window gaps, insulation spaces | ≈ 0.025 | Very poor conductor; good insulator. |
| Wood | Handles, building materials | ≈ 0.1–0.2 | Poor conductor; safe for pan handles. |
| Liquid water | Heating, cooling, climate | ≈ 0.6 | Better than air, still much worse than metals. |
| Glass | Windows, cookware lids | ≈ 1 | Poor–moderate conductor. |
| Stainless steel | Cookware, structures | ≈ 15 | Decent conductor, but not as high as copper. |
| Copper | Cookware, heat sinks, wiring | ≈ 400 | Excellent conductor; very small Rth. |
4. How conduction works in materials
In solids, heat is transferred mainly by particle collisions and, in metals, by free electrons.
- Metals: have many free electrons that move through the lattice, carrying energy quickly from hot regions to cold regions. This gives metals high k values.
- Insulators (wood, plastic, foam): have no free electrons and energy is transferred only through vibrations of atoms and molecules, which is slower.
- Liquids and gases: conduction is weaker, and heat transfer often relies more on convection (bulk motion of the fluid).
The temperature gradient dT/dx in Q̇ = −k A (dT/dx) tells us how quickly temperature changes with position. A steep gradient (large dT/dx) gives a larger rate of heat flow.
5. Real-life examples
(a) Metal spoon in hot soup
If one end of a metal spoon is in hot soup, the other end quickly becomes hot. The spoon has high thermal conductivity k, and the temperature difference between its ends produces a significant Q/t. That’s why the handle can become too hot to touch.
(b) Wooden handle vs metal handle
A cooking pan often has a metal body and a wooden or plastic handle. The metal part conducts heat rapidly (large Q/t), but the handle material has small k and large Rth, so the heat flow into your hand is small, keeping it comfortable to hold.
(c) Heat loss through house walls
In winter, heat flows from the warm interior to the cold outside. Using thick layers (large L) of low-k insulation decreases Q/t, reducing heating costs. Engineers use the conduction equation to size insulation and estimate energy use.
(d) Computer heat sinks
Computer processors generate heat that must be carried away to prevent overheating. Copper or aluminum heat sinks with large surface area A and high k conduct heat away from the chip quickly, then fins and fans remove that heat to the air.
6. Practice numerical problems (with solutions)
Problem 1: Heat flow through a wall
Question:
A brick wall is 0.20 m thick and has area 10 m². The inside surface is at 25 °C and the outside surface is at 5 °C. Take the thermal conductivity of brick as k = 0.7 W/m·K. Find the rate of heat flow through the wall (steady state).
Solution:
L = 0.20 m, A = 10 m², ΔT = 25 − 5 = 20 K.
Q/t = k A ΔT / L = (0.7)(10)(20) / 0.20 = 700 W.
Answer: 700 W.
Problem 2: Time to transfer a given amount of heat
Question:
A metal plate of area 0.5 m² and thickness 0.01 m has thermal conductivity k = 200 W/m·K. One side is kept at 100 °C and the other at 40 °C. How long will it take for 60,000 J of heat to flow through the plate?
Solution:
L = 0.01 m, A = 0.5 m², ΔT = 100 − 40 = 60 K.
Q/t = k A ΔT / L = (200)(0.5)(60) / 0.01 = 600,000 W.
t = Q / (Q/t) = 60,000 / 600,000 = 0.10 s.
Answer: 0.10 s.
Problem 3: Comparing two materials
Question:
Two rods of equal length L and equal cross-sectional area A are placed side-by-side between the same hot and cold reservoirs. Rod X is copper (k = 400 W/m·K), and rod Y is stainless steel (k = 15 W/m·K). Find the ratio of their heat flow rates (Q/t)X : (Q/t)Y.
Solution:
(Q/t)X / (Q/t)Y = kX / kY = 400 / 15 ≈ 26.7.
Answer: Copper conducts heat about 27 times faster than stainless steel.
7. Practice questions
- Q1. The SI unit of the rate of flow of heat (Q/t) is:
(A) J
(B) W
(C) J/K
(D) W/m·KCorrect answer: (B) W
- Q2. In the equation Q/t = k A (Thot − Tcold) / L, which change will decrease the rate of heat flow?
(A) Increasing the area A
(B) Increasing the temperature difference (Thot − Tcold)
(C) Increasing the thickness L
(D) Increasing the thermal conductivity kCorrect answer: (C) Increasing the thickness L
- Q3. The minus sign in Fourier’s law Q̇ = −k A (dT/dx) indicates that:
(A) Heat flows from cold to hot regions
(B) Heat flows from hot to cold regions
(C) The equation is only approximate
(D) Heat flow does not depend on temperature gradientCorrect answer: (B) Heat flows from hot to cold regions
- Q4. Which material below would have the largest thermal resistance Rth for the same area and thickness?
(A) Copper (k ≈ 400 W/m·K)
(B) Stainless steel (k ≈ 15 W/m·K)
(C) Wood (k ≈ 0.15 W/m·K)
(D) Glass (k ≈ 1 W/m·K)Correct answer: (C) Wood (k ≈ 0.15 W/m·K)
- Q5. For a given material and thickness, doubling the area A will:
(A) Halve the rate of heat flow
(B) Double the rate of heat flow
(C) Not change the rate of heat flow
(D) Reduce the rate of heat flow to one-quarterCorrect answer: (B) Double the rate of heat flow
8. Summary & conclusion
- The rate of flow of heat by conduction is described by Q̇ = −k A (dT/dx).
- For a uniform slab, Q/t = k A (Thot − Tcold) / L is widely used in problems.
- Thermal conductivity k varies strongly between materials; metals have high k, insulators have low k.
- Thermal resistance Rth = L/(kA) helps compare how easily heat flows through different layers.
- Real systems like walls, cookware, and electronics are designed using these ideas to control temperature and energy use.
- Understanding both the equation of rate of flow of heat and typical material values makes it easier to solve numerical questions and interpret real-world situations.
Heat conduction links microscopic particle motion to macroscopic energy flow. These equations let us estimate how quickly heat will move and help design safer, more efficient devices—from buildings and refrigerators to electronics and engines.
9. Frequently asked questions
1. What is the basic equation for the rate of flow of heat by conduction?
The basic equation is Q̇ = −k A (dT/dx), where Q̇ is the rate of heat flow, k is thermal conductivity, A is the cross-sectional area, and dT/dx is the temperature gradient. In many problems, it is used in the simplified slab form Q/t = k A (Thot − Tcold) / L.
2. Why is there a minus sign in Fourier’s law?
The minus sign in Q̇ = −k A (dT/dx) indicates the direction of heat flow. Heat flows from higher temperature to lower temperature, so the heat flow vector points opposite to the direction of increasing temperature. Mathematically, if temperature decreases with x (dT/dx < 0), then Q̇ is positive in the direction of heat flow.
3. What are typical units for each symbol in the conduction equation?
Common SI units are: Q in joules (J), t in seconds (s), Q/t in watts (W), k in W/m·K, A in m², L in m, and temperature difference (Thot − Tcold) in kelvins (K) or degrees Celsius (°C). With these units, the equation Q/t = k A ΔT / L is dimensionally consistent.
4. How is thermal resistance related to rate of heat flow?
Thermal resistance is defined as Rth = L/(kA). Using this, the rate of heat flow can be written as Q/t = ΔT / Rth. A larger Rth means a smaller Q/t for a given temperature difference, so materials with low k or large thickness act as better insulators.
5. When can I assume steady-state conditions in conduction problems?
Steady-state means that temperatures at each point in the material do not change with time (∂T/∂t = 0). You can usually assume steady-state in problems where the system has been held at constant boundary temperatures for a long enough time, so that the temperature profile no longer changes and the rate of heat flow Q/t is constant.
