1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

What would happen if an object of 689 degrees was in a room

  1. Oct 5, 2015 #1
    I was just thinking,

    If suddenly an object goes about 689 degrees farenheit (365 degrees celsius) in a room, about which perimeter would human life be impossible ?
  2. jcsd
  3. Oct 5, 2015 #2
    In case you need more elements, we will assume that the room is at 59 degrees farenheit (15 degrees celsius).
  4. Oct 5, 2015 #3
    And let's say the object is a cube of about 5,41 foot tall (1m65).
  5. Oct 5, 2015 #4


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    I'm sure the old cast iron pot belly stove/heater in my school classroom many, many yrs ago, got that hot and hotter
    it used to roar something dreadful and glow bright red to orange .... used to scare the hell out of me
    as a kid or around 10 yrs old

    similar to this just a bit taller and fatter

    My soldering iron gets to 450C and doesn't even glow red

  6. Oct 5, 2015 #5
    Yes but what about a bigger object ?
  7. Oct 5, 2015 #6


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    the pot belly was within close range to your measurements
    us kids were sitting within 2-3 metres of it
  8. Oct 5, 2015 #7
    And what about something like 2620 degrees Celsius ?
  9. Oct 5, 2015 #8


    User Avatar
    Gold Member

    What is your point with all this? What is it you are really trying to find out. Is there a point at which a large enough, hot enough object would make human life in a room impossible? Sure, but so what?
  10. Oct 5, 2015 #9
    Well it's pretty much a question after some sentence heard, which was something like "I produce 934% more heat when with you", so I just wandered what would actually happen if such an object just get that hot suddenly in a room, and about what distance would it take to be protected, out of complete curiosity. This is why the question was asked in the "General Discussion" forum.
  11. Oct 5, 2015 #10
    That's a bit too high for anyone to sit close to.
  12. Oct 5, 2015 #11
    The thing is I started thinking the normal body temperature is 37 degrees Celsius. So if you take 934 percent more it goes up to 365 degrees Celsius, which explains my first question.
    Then it occurred to me that in a physic way it would be more legitimate in Kelvin. So if we start at 310,15 K (37 degrees Celsius), and add 934% we get to approximatively 2890 (2620 degrees Celsius), which is why I asked the second question.
  13. Oct 5, 2015 #12
    Yes I know, the real question is how can we find out the good distance not to be harmed ?
  14. Oct 5, 2015 #13
    According to this:


    steel glowing orange is between 1708 and 1908 degrees F.
  15. Oct 5, 2015 #14
    I had one of those huge cast iron fireplaces too, our cat would sleep directly underneath it.

    A cast iron pan easily gets 400 degrees and not only can you put your hand almost directly above it with no danger, you can actually grab the handle, which is also cast iron and attached to the pan.

    It would have to do with how small the room is and how well insulated it was.
  16. Oct 5, 2015 #15


    User Avatar

    Staff: Mentor

    No, radiative heat transfer scales to the fourth power, so the temperature to output 4x more heat is much, much lower than that....though it is complicated for relatively cool objects since they are convection, not radiation dominated.

    [Moved to general physics]
  17. Oct 5, 2015 #16


    User Avatar

    You do have to do it in Kelvin, since that's the only way it makes sense. However, as russ said, radiative heat transfer scales as T4, so to output 9.34 times as much heat, the object would need to be much less than 9.34 times as warm. A human outputs something like 100 watts of heat, so something outputting 934% the heat would be outputting in the neighborhood of a kilowatt. This isn't going to make a room unlivable, nor would it be difficult to stand near - it would be about the same output as a smallish space heater.
  18. Oct 5, 2015 #17


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Perhaps look up "power emitted by a black body"
  19. Oct 5, 2015 #18
    If the "object" is a human being the first thing to happen is that he/she will die and stop producing any heat pretty soon.
  20. Oct 6, 2015 #19
    Depends on the time spent, note that a longer duration inside an isolated room with an object of that temperature would cause dehydration. Also if the object is warm enough it can cause damage to the skin. Because two objects tend to mix their temperature, laws of thermodynamics. For example let's say room temp is 15 and the object is 900 whatever unit for both. So for some time the room temp. would have gotten to be 885 units I think. Basically you get the idea, time would be a decisive factor.
  21. Oct 6, 2015 #20
    To get an answer it just needs to be more specific: Say the room is 4x4x3meters (3 tall) and the object has a one meter square surface facing you and it's in the middle of the room so you are 2 meters away (at best!). We want to know how hot this surface need to be to incinerate you. I'll assume lighting your clothes on fire within 1 second is a sufficient condition. So I'll say your clothing temperature needs to go to 500F or 260C (Fahrenheit 451?). Let's also assume the object has perfect emissivity (e=1, perfect blackbody) and that the your clothing has e=0.50 - half the radiation is absorbed. Further let's say you block 1 m^2 of this radiation. Finally we need the specific heat capacity of the clothing (how fast will it heat up). Let's say about 500 J/C for 1 kilogram of clothing. Starting at room temp we need 130kJ (about a shot glass of gasoline!). Energy from the "object is E = 5.67x10^-8*T^4*A where A we said is 1 and the constant is Boltzmann's in SI. Now if this were a point source we could use 1/r^2 but in this case we need to estimate the fraction area intercepted by our presence. We can assume that total area is the walls of our half of the room (not quite true but good enough) . So the "E" spreads over half the room walls = 2x4x4+4x3 =44m2 (1/2 the area of a sphere 2 meters diameter =50m2). So our 1 meter squared self will absorb 1/44th of E. We want 1/44th of E to equal 130kJ OR 130,000/(5.67x10^-8) = T^4. I get 1230K = 930 C or 1700 F.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook