Conduction, Convection, and Radiation.

Key Idea 🔑

One of the most important points to remember when dealing with any type of heat transfer problems is that thermal energy refers to the heat energy generated by molecular movement & collisions within an object. In other words, whenever molecules bump into each other, there's going to be thermal energy involved! 🔥

Types of Thermal Conductivity 💻

Thermal conductivity might feel like a much more loaded word than just “heat transfer,” but it’s actually a pretty straightforward and simple concept once you get the hang of it. 👍

Think about how one surface often feels colder than another when you touch it—take, for example, stepping on a tile floor after walking on carpet. The difference in perceived temperatures of the surfaces is because different materials transfer heat at different rates. 🥶

Generally, good conductors of electricity (metals like copper and aluminum) are also good heat conductors. In the same way, insulators (wood, rubber, etc.) are poor heat conductors. ⚓

When two objects at different temperatures collide, an energy transfer from the hotter to the colder occurs. This transfer of heat between two objects that come in contact is what we call thermal conduction. 🚶‍♂️🚶‍♀️

💡 Note: Generally, it's better to think of coldness as an absence of heat energy rather than "coldness" entering a place or object.

1️⃣ Conduction

As mentioned above, conduction is the transferral of heat through direct contact. It occurs in solids through molecular collisions. 💥

Image courtesy of Wikimedia

2️⃣ Convection

Convection is a mode of heat transfer that involves the movement of molecules and is most common in fluids, and it occurs naturally as well, controlled by gravity. 🌊

3️⃣ Radiation

Radiation is a little different: it occurs through electromagnetic waves in empty space instead! So, think of something like the heat radiating off of a hot object.

Image courtesy of Wikipedia

It doesn’t require any physical contact to occur either and occurs naturally at slight temperature increases. Radiation is actually emitted by all objects that exist at a temperature greater than 0 Kelvin (absolute zero); it’s just at too small of a frequency to be felt most of the time. 0️⃣

Equation for Thermal Conductivity ➗

There are four factors that affect the rate at which heat is conducted through a material:

• k: the thermal conductivity constant of the material

• k is larger for materials that transfer heat well, and smaller for materials that transfer heat poorly

• A: the cross-sectional area of the material that is transferring heat

• The number of molecular collisions raises with the increased area of both objects that are in contact

• Delta-T: the difference in temperature between both objects

• d: the thickness of the material

• The thicker the material, the more time it takes to transfer the same amount of heat!

The effects of these 4 factors can be seen in this equation for the rate of thermal conduction:

(The letter Q represents the amount of heat transferred in time t. The entire left side of the equation represents the number of joules (J) of heat energy transferred through the material per second. The rate of conductivity has units of joules/second = watts.) ⚡

💡 Note: 

• There is a direct relationship between each of k, A, and delta-T with the heat flow Q/t. This is shown in the equation by the numerator having all of these quantities on the right-hand side. When any of them increases, so does the amount of Q/t. 

• In the same way, there is an inverse relationship between d and the flowrate. This is shown by d being in the denominator in the equation. As the distance between the objects increases, the flow rate of heat Q/t decreases.

Practice Problem 📝

A woman wants to replace her house’s front window, but she doesn’t want to change how much heat the window retains. Which of the following changes could be made to the window that would leave it with the same thermal conductivity rate as before? The original window has an area of 2A and a thickness of 4d and is made of glass with thermal conductivity constant k.

A. double the area, cut the thickness in half, cut k in half

B. quadruple the area, double the thickness, cut k in half

C. double the area, double the thickness, quadruple k

D. cut the area in half, double the thickness, double k

😎 Solution 

For this problem, we'll use the rate of thermal conduction listed above. To find out which changes would leave the rate of thermal conduction Q/t unchanged, simply multiply each variable by its given constant and solve.

CORRECT ANSWER: B