Video: Determining the Required Temperature to Produce a Particular Radiated Power

An infrared heater has a surface area of 0.050 m² and an emissivity of 0.84. What temperature must it run at if the required power is 360 W? Neglect the temperature of the environment.

02:49

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.

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