Read this article to learn how to determine the rate at which heat conducts through a material.

What is thermal conduction?

Walking on bathroom tile in winter is annoying since it feels so much colder than the carpet. This is interesting, since the carpet and tile are usually both at the same temperature (i.e. the temperature of the interior of the house). The different sensations we feel is explained by the fact that different materials transfer heat at different rates. Tile and stone conduct heat more rapidly than carpet and fabrics, so tile and stone feel colder in winter since they transfer heat out of your foot faster than the carpet does.
People usually think of the sensation of cold as "coldness entering the body", but feeling cold is better thought of as "heat energy leaving the body".
In general, good conductors of electricity (metals like copper, aluminum, gold, and silver) are also good heat conductors, whereas insulators of electricity (wood, plastic, and rubber) are poor heat conductors. The figure below shows molecules in two bodies at different temperatures. The (average) kinetic energy of a molecule in the hot body is higher than in the colder body. If two molecules collide, an energy transfer from the hot to the cold molecule occurs. The cumulative effect from all collisions results in a net flux of heat from the hot body to the colder body. We call this transfer of heat between two objects in contact thermal conduction.

Image: The molecules in two bodies at different temperatures have different average kinetic energies. Collisions occurring at the contact surface tend to transfer energy from high-temperature regions to low-temperature regions. (Image Credit: Openstax College Physics)

What's the equation for the rate of thermal conduction?

There are four factors (kk, AA, ΔT\Delta T, dd) that affect the rate at which heat is conducted through a material. These four factors are included in the equation below that was deduced from and is confirmed by experiments.
Qt=kAΔTd\Large \dfrac{Q}{t}=\dfrac{kA\Delta T}{d}
The letter QQ represents the amount of heat transferred in a time tt, kk is the thermal conductivity constant for the material, AA is the cross sectional area of the material transferring heat, ΔT\Delta T is the difference in temperature between one side of the material and the other, and dd is the thickness of the material. These factors can be seen visually in the diagram below.
Image: Heat conduction occurs through any material, represented here by a rectangular bar, whether window glass or walrus blubber. (Image Credit: Openstax College Physics)

What does each term represent in the thermal conduction equation?

There's a lot to digest in the equation for thermal conduction Qt=kAΔTd\dfrac{Q}{t}=\dfrac{kA\Delta T}{d}. Let's look at what each factor means individually below.
Qt\dfrac{Q}{t}: The factor on the left hand side of the equation (Qt)(\dfrac{Q}{t}) represents the number of joules\text{joules} of heat energy transferred through the material per second\text{second}. This means the quantity Qt\dfrac{Q}{t} has units of joulessecond=watts\dfrac{\text{joules}}{\text{second}}=\text{watts}.
kk: The factor kk is called the thermal conductivity constant. The thermal conductivity constant kk is larger for materials that transfer heat well (like metal and stone), and kk is small for materials that transfer heat poorly (like air and wood).
For a given material, kk is a constant. But each material has its own different constant kk value. For examples of thermal conduction constants check out the chart below.
Materialk in units of JsmoC\dfrac{\text{J}}{\text{s}\cdot \text{m} \cdot ^o\text{C}}
silver420
copper390
gold318
glass0.84
water0.6
wool0.04
air0.023
styrofoam0.010
(Openstax College Physics)
ΔT\Delta T: The heat flow is proportional to the temperature difference ΔT=ThotTcold\Delta T=T_{hot}-T_{cold} between one end of the conducting material and the other end. Therefore, you will get a more severe burn from boiling water than from hot tap water. Conversely, if the temperatures are the same, the net heat transfer rate falls to zero, and equilibrium is achieved.
AA: Owing to the fact that the number of collisions increases with increasing area, heat conduction depends on the cross-sectional area AA. If you touch a cold wall with your palm, your hand cools faster than if you just touch it with your fingertip.
dd: A third factor in the mechanism of conduction is the thickness dd of the material through which heat transfers. The figure above shows a slab of material with different temperatures on either side. Suppose that T2T_2 is greater than T1T_1, so that heat is transferred from left to right. Heat transfer from the left side to the right side is accomplished by a series of molecular collisions. The thicker the material, the more time it takes to transfer the same amount of heat. This model explains why thick clothing is warmer than thin clothing in winters, and why Arctic mammals protect themselves with thick blubber.

Why do metals feel both colder in the winter, and hotter in the summer?

Materials with a high thermal conductivity constant kk (like metals and stones) will conduct heat well both ways; into or out of the material. So if your skin comes into contact with metal that is colder than your skin temperature, the metal can rapidly transfer heat energy out of your hand, making the metal feel particularly cold. Similarly, if the metal is hotter than your skin temperature, the metal can rapidly transfer heat energy into your hand, making the metal feel particularly hot.
This is why concrete will feel especially cold to our bare feet in winter (the concrete transfers heat out of our feet rapidly), and especially hot to our bare feet in summer (the concrete transfers heat into our feet rapidly).

What do solved examples involving thermal conduction look like?

Example 1: Window makeover

A person wants to replace the window on her house, but she doesn't want her heating and cooling bills to change. The original window on the wall of the house has area AA and thickness dd, and is made out of glass that has a thermal conduction constant kk.
Which one of the following changes could be made to the window that would leave the rate of thermal conduction the same as the original window? (Select one)
Choose 1 answer:
Choose 1 answer:
The formula for the rate of thermal conduction is Qt=kAΔTd\dfrac{Q}{t}=\dfrac{kA\Delta T}{d}. To find out which changes to the window would leave the rate of thermal conduction unchanged, we need to figure out which changes leave the right hand side of the formula unchanged.
If we double the area, double the thickness, and quadruple the k constant the rate of thermal conduction would become Qt=(4k)(2A)ΔT(2d)=4kAΔTd\dfrac{Q}{t}=\dfrac{(4k)(2A)\Delta T}{(2d)}=4\dfrac{kA\Delta T}{d}, so this would quadruple the rate of thermal conduction.
If we double the area, cut the thickness in half, cut the k constant in half the rate of thermal conduction would become Qt=(k/2)(2A)ΔT(d/2)=2kAΔTd\dfrac{Q}{t}=\dfrac{(k/2)(2A)\Delta T}{(d/2)}=2\dfrac{kA\Delta T}{d}, so this would double the rate of thermal conduction.
If we cut the area in half, cut the thickness in half, and double the k constant the rate of thermal conduction would become Qt=(2k)(A/2)ΔT(d/2)=2kAΔTd\dfrac{Q}{t}=\dfrac{(2k)(A/2)\Delta T}{(d/2)}=2\dfrac{kA\Delta T}{d}, so this doubles rate of thermal conduction.
If we quadruple the area, double the thickness, cut the k constant in half the rate of thermal conduction would become Qt=(k/2)(4A)ΔT(2d)=kAΔTd\dfrac{Q}{t}=\dfrac{(k/2)(4A)\Delta T}{(2d)}=\dfrac{kA\Delta T}{d}, so this would leave the rate of thermal conduction the same.

Example 2: Window heat loss

A single-paned window in your house is 0.65 m0.65\text{ m} wide, 1.25 m1.25\text{ m} tall, and has a thickness of 2 cm2\text{ cm}. The glass has a thermal conduction constant of 0.84JsmoC0.84 \dfrac{\text{J}}{\text{s}\cdot \text{m} \cdot ^o\text{C}}. Assume the outside temperature of the glass is a constant 5o C5^o\text{ C} and the inside temperature of the glass is a constant 20o C20^o\text{ C}.
How many joules\text{joules} of heat are transferred out of the window in one hour?
Solution:
Qt=kAΔTd(start with the formula for the rate of thermal conduction)\dfrac{Q}{t}=\dfrac{kA\Delta T}{d} \quad {\text{(start with the formula for the rate of thermal conduction)}}
Q=tkAΔTd(multiply both sides by  to isolate )tQQ=\dfrac{tkA\Delta T }{d} \quad {\text{(multiply both sides by $t$ to isolate $Q$)}}
Q=(3600 s)kAΔTd(the time interval is , which is )1 hour3600 secondsQ=\dfrac{(3600\text{ s})kA\Delta T }{d} \quad {\text{(the time interval is $1 \text{ hour}$, which is $3600 \text{ seconds}$)}}
Q=(3600 s)(0.84JsmoC)AΔTd(plug in the  value for the glass)kQ=\dfrac{(3600\text{ s})(0.84 \dfrac{\text{J}}{\text{s}\cdot \text{m} \cdot ^o\text{C}})A\Delta T }{d} \quad {\text{(plug in the $k$ value for the glass)}}
Q=(3600 s)(0.84JsmoC)(0.8125 m2)ΔTd(the area is )height×width=0.65 m×1.25 m=0.8125 m2Q=\dfrac{(3600\text{ s})(0.84 \dfrac{\text{J}}{\text{s}\cdot \text{m} \cdot ^o\text{C}})(0.8125 \text{ m}^2)\Delta T }{d} \quad {\text{(the area is $\text{height} \times \text{width}=0.65\text{ m}\times1.25\text{ m}=0.8125\text{ m}^2$)}}
Q=(3600 s)(0.84JsmoC)(0.8125 m2)(15oC)d()ΔT=ThotTcold=20oC5oC=15oCQ=\dfrac{(3600\text{ s})(0.84 \dfrac{\text{J}}{\text{s}\cdot \text{m} \cdot ^o\text{C}})(0.8125 \text{ m}^2)(15^o\text{C}) }{d} \quad {\text{($\Delta T=T_{hot}-T_{cold}=20^o\text{C}-5^o\text{C}=15^o\text{C}$)}}
Q=(3600 s)(0.84JsmoC)(0.8125 m2)(15oC)0.02 m(the thickness  must be in meters, )d2 cm=0.02 mQ=\dfrac{(3600\text{ s})(0.84 \dfrac{\text{J}}{\text{s}\cdot \text{m} \cdot ^o\text{C}})(0.8125 \text{ m}^2)(15^o\text{C})}{0.02\text { m}} \quad {\text{(the thickness $d$ must be in meters, $2\text{ cm}=0.02\text{ m}$)}}
Q=1.84×106 J(calculate and celebrate)Q= 1.84 \times 10^6 \text{ J}\quad {\text{(calculate and celebrate)}}\quad
1.84×106 J1.84 \times 10^6 \text{ J} of thermal energy is enough to melt about 0.8 kg0.8 \text{ kg} of ice. That would be an ice cube with side lengths of about 10 cm10\text{ cm}.
Because so much heat can be transferred through windows, people developed double-paned windows that have a layer of air sandwiched in between two panes of glass. Since air conducts heat less than glass, this extra layer of air reduces the rate of heat transfer through a window.
This article was adapted from the following article:
  1. "Conduction" from Openstax College Physics. Download the original article free at http://cnx.org/contents/031da8d3-b525-429c-80cf-6c8ed997733a@9.4:105/Conduction
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