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There is an 8x8x8 room on the moon

  1. Apr 26, 2004 #1
    I was asked the following question by a friend and don't know how to help him, or where to even start. I'll admit, I'm a lawyer and its been several years since I've taken a physics class. If anyone can help I'd appreciate it.

    There is an 8x8x8 room on the moon; in the room is a refrigerator that is plugged in but I'm not told if it is working or not. However, I presume it is working since there is a 60 watt light bulb in it that is on. There is no air in the room; it is like a vacuum. I was asked to figure out what the temperature of the room is and to figure out as the bulb stays on whether the temp of the room increases or decreases.

    Thanks for the help.
  2. jcsd
  3. Apr 26, 2004 #2
    I cannot answer your first question, but I believe that heat fluxuation is defined by the accumulative motion or non-motion of particles in the room - therefore I am pretty sure that the area surrounding the lightbulb is subject to the light's acceleration of particles. So, I would think that the temperature would increase, if the light bulb was able to heat in the first place, considering the moon is relatively much colder than the heat necessary to glow a fuse.
  4. Apr 26, 2004 #3
    *waits for physics professor to beat him mercilessly
  5. Apr 26, 2004 #4
    What exactly is the point of attempting to cool a vacuum again? Or, in other words, what exactly is the temperature of a vacuum?

  6. Apr 27, 2004 #5


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    In another recent thread it was discussed how the space shuttle opens the bay doors while in orbit to allow the heat from the electronics to radiate out of the craft to avoid heat buildup.

    In this case the bulb would heat the refridgerator walls, and after that heat makes it way past the insulation.....

  7. Apr 27, 2004 #6


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    Wherever the light can reach, the room will be heated by the bulb. Energy is being brought in from somewhere through the plug in the fridge, and radiated out to the room by the bulb. In a vacuum, there is no air to "spread the heat out", so whatever is directly struck by the light will be warmed, but the parts of the room that are in the shade will receive none of this warmth.

    Perhaps more important than wether or not the fridge is working; is the door open?
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