 # Thermal Conductivity

Understanding the thermal conductivity of materials lets you become a better cook. It also allows you to choose the right clothes to wear to avoid being too hot (or cold). Plus, it helps you pick the right home insulating materials, which can save you as much as 20% on heating and cooling costs!

Below, we answer your important questions about thermal conductivity and we'll even share a few tricks to solving conductivity-related formulas and problems, so, keep reading!

## What is thermal conductivity (in 100 words)?

Thermal conductivity refers to a material's intrinsic property and ability to conduct heat. It's one of the three ways of transferring heat. The other two are convection and radiation.

Heat conductivity is a measurement of heat quantity, often shown or written as watts per meter-kelvin units. Through conduction, transfer of heat takes place through direct, physical contact.

Temperature difference and the material's surface area are two factors that affect conductivity. But you also need to consider the distance and time the heat needs to travel, as well as the type of materials involved (and their properties).

## Which factors affect conductivity?

To arrive at a material's conductivity rate, you need to know the value of the following variables:

• Temperature difference (ΔT)
• Cross-sectional area (A)
• Distance (L)
• Time (t)

To make things easier, let's use a rectangular metal bar for example.

### Temperature Difference (ΔT)

Temperature difference refers to the difference in temperatures within the materials involved. In this case, the difference between one end of the bar and the other.

The bigger the temperature difference between these areas, the higher the thermal energy transfer rate. That means greater amounts of heat transferred.

### Cross-Sectional Area (A)

The bigger the material, the greater the amount of heat needed to heat it up. You also need to account for exposure to open air. The larger the exposed area, the higher the possibility of losing heat along the way.

### Distance or Length (l)

This refers to the distance that heat needs to travel. As such, the shorter the metal bar, the more heat it'll allow through.

### Time (t)

The amount of time (t) wherein heat travels affects the amount of heat that will transfer (Q). The more time you let pass, the more heat that will make it through the bar.

### Material Properties (k)

The materials' physical properties (expressed as "k") also play a part in the formula for rate of conductivity. Most common metals have good thermal conductivity, while materials such as paper, cloth, and air have poor heat conductivity.

When calculating thermal conductivity, k is one value you'll have no problems getting. That's because physics has already given us the thermal conductivity value for many common materials.

You may also encounter the term "temperature gradient" in some formulas. This refers to the physical measurement describing the direction and speed (rate) of changes in the temperature of a location.

## Whats the formula for conductivity rate?

Now that you know the factors, variables, and givens, you can already find out how much heat transfer occurs through conduction. Here's one of the formulas you can use:

#### Q = k· A · ΔT · t / l

Wherein:

• Q refers to the value (amount) of heat that conduction transfers
• The "k" is the given thermal conductivity of the material (such as copper)
• A is the surface area
• ΔT is the temperature difference throughout the material
• The "t" is the given amount of time for heat to travel
• The "l" is the length or distance that heat needs to travel

That's a lot to consider, right? Don't worry. Below is a solved equation sample for heat loss through a window.

## What's an example of a solved equation?

Say you want to know how many joules (J) of heat exits a window of a house every hour, in seconds.

The single-paned window has a width of 0.75 meters, a height of 1 meter, and a thickness of 1.5 centimeters.

The glass' thermal conduction constant is 0.96 J/s·cm·°C. Outside, the temperature is constant at 10°C, while inside the glass is constant at 22°C.

Let's convert everything we need first.

#### Area (A)

= Width x height

= 0.75 meters x 1 meter

= 0.75m2

#### Temperature Difference (ΔT)

= Hot temperature - cold temperature

= 22°C - 10°C

= 12°C

#### Time

1 hour = 3,600 seconds (s)

#### Distance (l) in meters

= 1.5 cm / 100 (100 centimeters in every meter)

= 0.015 m

### Solution

Q = k·A·ΔT·t / l

Q = (0.96 J/scmC) (0.75 m2) (12°C) (3,600 seconds) / 0.015 m

Q = 2.07 x 106 J

This means that you lose 2.07 x 106 J of thermal heat through that window.

## What is the heat conductivity of common materials?

You don't need to memorize the heat conductivity of all materials known to man. But it still pays to know those for some of the most common materials.

### Air

The thermal conductivity of air at 0 °Celsius is 0.024 W/m, °Kelvin. You can also express it as 0.000057 (cal/sec)/(cm2, °C/cm).

### Water

The thermal conductivity of water at 20 °Celsius is 0.0014 cal/sec/cm2°C.

### Copper

Pure copper's in imperial units is 223 BTU/h/ft °F. BTU stands for British Thermal Units, h or hr for "hour," ft for foot, and °F for degree Fahrenheit.

In the International System of Units (SI), it's expressed as 386 W/m K. W stands for "Watt," m for "meter" and "K" for Kelvin.