### Video Transcript

An infrared heater has a surface
area of 0.050 meters squared and an emissivity of 0.84. What temperature must it run at if
the required power is 360 watts? Neglect the temperature of the
environment.

So in this question, we’ve got an
infrared heater which has a surface area of 0.050 meters squared and an emissivity
of 0.84. We need to find the temperature
that it runs at if the required power is 360 watts. And we can neglect the temperature
of the environment. This last sentence is telling us
basically that we only need to consider the heater in isolation. In other words, it’s making life a
little bit easier for us. We don’t have to think about
anything, except for the heater.

So let’s start labelling all the
quantities. We’ll call the area of the heater
𝐴, which is 0.050 meters squared. The emissivity we’ll call 𝜀, which
is 0.84. The power we’ll call 𝑃, which is
360 watts. And we’re trying to find out the
temperature, which we’ll call 𝑇. So we need to find the relationship
that links together 𝐴, 𝜀, 𝑃, and 𝑇.

The relationship we’re looking for
is known as the Stefan–Boltzmann law. It tells us that the power per unit
area, which is the left-hand side, is equal to the emissivity multiplied by the
Stefan–Boltzmann constant, which is just a number, multiplied by the temperature to
the fourth power. So it relates all of the quantities
that we’ve just mentioned above.

And by the way, as an aside, the
left-hand side of the equation is often known as the power flux because it’s the
amount of power per unit area or the amount of power being radiated through a unit
area. But for now, that’s not
important. We just need to go on to solve for
the temperature.

So if we divide both sides of the
equation by 𝜀𝜎, then we find that 𝑃 over 𝐴𝜀𝜎 is equal to 𝑇 to the power of
four. At this point, we can take the
fourth root of both sides of the equation. And of course, the fourth root
cancels with the fourth power, leaving us with just the temperature on the
right-hand side. At this point, all that’s left for
us to do is to substitute in the values.

So we know that 𝑇 is equal to the
power divided by the area multiplied by the emissivity multiplied by the
Stefan–Boltzmann constant, which happens to be 5.67 times 10 to the power of
negative eight watts per meter squared kelvin to the fourth. And we need to take the fourth root
of this whole expression. So evaluating this gives us a
temperature of 623.54 dot dot dot kelvin.

But of course, out of the values
that we’ve been given at the question, the lowest number of significant figures
given to us is two. So we have to round our answer to
two significant figures as well. And so our final answer is that the
temperature of the heater must be 620 kelvin.