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Heating/cooling a volume of air.

  1. Feb 18, 2008 #1
    I have a box that is 5'w x 5'l x 3.3'h and the air temperature within the box is 60F. If I have a device that can heat the air in the box to 80F in 7 minutes, then what formula could I use to determine how long it would take that same device to heat a box exactly 3 times the size of the first box (15'w x 15'l x 10'h) from 60F to 80F?

    I know its not going to be a linear (7min x 3 = 21 min) equation because larger volumes of gas or liquid take longer to cool or heat than smaller volumes.

    In addition, could this same formula be applied to determine how long it would take either boxes to heat from 60F to 90F, or possibly to cool from 90F to 50F? I am hoping to find a general formula that can be applied to a variety of heating/cooling scenarios.

    Thank you.
  2. jcsd
  3. Feb 18, 2008 #2


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    You're going to need a little more information to apply the right heat transfer formula. What is the current dominant mechanism when heating the box? Natural convection? Forced convection? Radiation?
  4. Feb 18, 2008 #3


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    In simple terms (assuming the air mixes and is well insulated) it is linear.
    But it is not 3x, the second box is not3 times the volume it is 27 times (3x3x3) the volume so will take 27times as long to heat the air.
    In the real world there are probably losses from the surfaces which are 9 times as large so it will take a bit longer than 27x.
  5. Feb 18, 2008 #4
    If the heater is at a much higher temperature of the box then the power transfered per unit time should be roughly constant. In which case you can assume

    [tex]\Delta T = C_p V_1 P_{heater} t_2 \Delta[/tex]

    [tex]\Delta T = C_p V_2 P_{heater} t_1 \Delta[/tex]
    [tex]C_p[/tex] is the specific heat
    [tex]V_1[/tex] is the volume of box 1
    [tex]V_1[/tex] is the volume of box 1
    [tex]\Delta T[/tex] is the change in temperature.
    [tex]t_1[/tex] is the time to heat box 1
    [tex]t_2[/tex] is the time to heat box 2
    [tex]P_{heater}[/tex] is the power of the heater.

    You cannot apply the same equation to cooling because you would need more information as the above poster mentioned. See:
    Last edited: Feb 18, 2008
  6. Feb 20, 2008 #5
    Some more info to assist the question:

    I have a fan that disperses air at 8 CFM, the air is heated to 80 F.

    Assuming a fully insulated box that is 5'x5'x3 = 75 Cubic Feet. Thus 75CF / 8CFM = 9.3 minutes to heat, but this basic equation doesn't take into account the existing temperature of the air in the box.

    The same equation should apply to a 15'x15'x10 box. 2250CF / 8 CFM = 4.7 hours, again, not taking into account the current temperature of the box.

    If the known temperature of either box is exactly 60F...this should be enough info to compute a more accurate amount of time to heat the box, right?
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