Change the chapter
Question
At what net rate does heat radiate from a $275 \textrm{ m}^2$ black roof on a night when the roof’s temperature is $30.0 ^\circ\textrm{C}$ and the surrounding temperature is $15.0 ^\circ\textrm{C}$? The emissivity of the roof is 0.900.
Question by OpenStax is licensed under CC BY 4.0.

$-2.18 \times 10^4 \textrm{ W}$

Solution Video

OpenStax College Physics for AP® Courses Solution, Chapter 14, Problem 55 (Problems & Exercises) (1:45)

Sign up to view this solution video!

Rating

No votes have been submitted yet.

Quiz Mode

Why is this button here? Quiz Mode is a chance to try solving the problem first on your own before viewing the solution. One of the following will probably happen:

  1. You get the answer. Congratulations! It feels good! There might still be more to learn, and you might enjoy comparing your problem solving approach to the best practices demonstrated in the solution video.
  2. You don't get the answer. This is OK! In fact it's awesome, despite the difficult feelings you might have about it. When you don't get the answer, your mind is ready for learning. Think about how much you really want the solution! Your mind will gobble it up when it sees it. Attempting the problem is like trying to assemble the pieces of a puzzle. If you don't get the answer, the gaps in the puzzle are questions that are ready and searching to be filled. This is an active process, where your mind is turned on - learning will happen!
If you wish to show the answer immediately without having to click "Reveal Answer", you may . Quiz Mode is disabled by default, but you can check the Enable Quiz Mode checkbox when editing your profile to re-enable it any time you want. College Physics Answers cares a lot about academic integrity. Quiz Mode is encouragement to use the solutions in a way that is most beneficial for your learning.

Calculator Screenshots

OpenStax College Physics, Chapter 14, Problem 55 (PE) calculator screenshot 1
OpenStax College Physics, Chapter 14, Problem 55 (PE) calculator screenshot 2
Video Transcript
This is College Physics Answers with Shaun Dychko. The roof has an area of 275 square meters and its temperature is 303.15 Kelvin after you add 273.15 to 30 degrees Celsius. And, the temperature of the surroundings at 15 degrees Celsius converts to 288.15 Kelvin, and of course always taking care of unit conversions in this step where you're writing down the data given in order to avoid having to think about it later when you're thinking about more important things like Algebra. And, the emissivity, we're told of this dark roof, is 0.9. So, this is the rate of heat transfer to an object due to radiation and it's the Stefan Boltzmann constant times the emissivity times the area times the temperature of the surroundings to the power four minus the temperature of the object to the power of four. So, we substitute in all the numbers. That's 5.67 times ten to the minus eight Joules per second per meters squared per Kelvin to the fourth. And then, times by emissivity of 0.90 times the area of 275 square meters times 288.15 Kelvin, temperature of the surroundings, to the power four minus 303.15 Kelvin, temperature of the roof, to the power four. And, this makes negative 2.18 times ten to the four watts. And so, the roof is gaining energy at this rate but since it's negative, it means in fact it is losing heat to the surroundings which is what we expect because the temperature of the roof is higher than the surroundings are. Thus, we expect it to be losing energy and in fact it is, and we can see that from this negative sign.