Converting watt into temperature (celcius)

Discussion in 'General Electronics Chat' started by rajat1684, Aug 28, 2013.

  1. rajat1684

    Thread Starter New Member

    Aug 26, 2013
    13
    0
    Thanks you all.

    Intersting..I think I can understand now why more power (watt) means more heat (KJ).

    Converting watt into temperature (celcius)

    Say 70% of 50 watt of Halogen Bulbs converts into heat (hypothetically)
    i.e. = 0.7*50 = 35 watt of heat

    Bulb is on for 5 minutes = 60*5 = 300 sec

    1 watt = 1 joule/ sec, thus 35*300 = 10500 joules = 105 KJ
    Specific Heat of Dry Air (Cv) = 0.716 KJ/kg.K
    Thus,
    105KJ/0.716KJ/kg.K = 146 Kg.K
    Density of Air = 1.3 Kg/m3

    Thus 146 (Kg.K)/1.3 (Kg/m3) = 112.3 K-m3

    This means temperature rise will be 112.3 K per m3 of dry air.

    Converting Kelvin to Celsius

    Celsius = Kelvin- 273
    Celcius = 112.3-273 = -160.7 (taking mode = 160 degree celcius)

    Is this correct that temperature rise will be 160 degree celcius?
     
  2. rajat1684

    Thread Starter New Member

    Aug 26, 2013
    13
    0
    Sorry because I'm not an electrical engineer any advice would be good for me
     
  3. wayneh

    Expert

    Sep 9, 2010
    12,958
    3,775
    A degree kelvin is the same size as a degree celsius, so no need to go any farther once you have °K.

    Your calculation is reasonable although I haven't checked every detail. One problem, though, is that both the density and the heat capacity are not constant, but are functions of temperature themselves. So the precise answer requires integral calculus.

    Depending on what happens to the light, it might end up heating the air as well.
     
  4. rajat1684

    Thread Starter New Member

    Aug 26, 2013
    13
    0
    Thanks wayneh :)
     
  5. rajat1684

    Thread Starter New Member

    Aug 26, 2013
    13
    0
    It just seems lot of rise in temp so just thought to share it in forum
     
  6. wayneh

    Expert

    Sep 9, 2010
    12,958
    3,775
    Oh, hey, you slipped a decimal in converting to kJ.
     
  7. GopherT

    AAC Fanatic!

    Nov 23, 2012
    6,837
    5,216
    Here...with errors corrected.

     
  8. PaulEngineer

    Member

    Dec 21, 2016
    143
    6
    Uuuuum the 10500 J is not 105KJ it is 10,5KJ ;)! 105KJ = 105000J
     
  9. hp1729

    Well-Known Member

    Nov 23, 2015
    2,309
    279
    Are there other factors? Does the housing for the light provide a heat sink? On a power transistor for example, I can be dissipating 20 Watts but with a large enough heat sink temperature does not rise considerably.
     
  10. wayneh

    Expert

    Sep 9, 2010
    12,958
    3,775
    Old thread, guys.
     
  11. PaulEngineer

    Member

    Dec 21, 2016
    143
    6
    Hi
    It is true! The heat sink is the only thing that protects a power transistor from its overheat! As better heat sink you will put (always with fan near) so big is the chance to rise the voltage and the current! The only thing that destroying the transistor, is the heat! If you will put a good coolant system, you can rise the voltage and current! I made a 500W amplifier that works great on 60V and 8A which makes 480W (but with some capacitors 6800uF/63V it gives 20 more volts) and all that thanks to the good coolant system!
     
Loading...