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?

 

Sorry because I'm not an electrical engineer any advice would be good for me

 

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.

 

Thanks wayneh

 

It just seems lot of rise in temp so just thought to share it in forum

 

Oh, hey, you slipped a decimal in converting to kJ.

 

Oh, hey, you slipped a decimal in converting to kJ.

Here...with errors corrected.

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 = 10.5 KJ
Specific Heat of Dry Air (Cv) = 0.716 KJ/kg.K
Thus,
10.5KJ/0.716KJ/kg.K = 14.6 Kg.K
Density of Air = 1.3 Kg/m3

Thus 14.6 (Kg.K)/1.3 (Kg/m3) = 11.23 K-m3

This means temperature rise will be 11.23 K per m3 of dry air.

Converting Kelvin to Celsius
11.23 C temperature rise.

 

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?

Uuuuum the 10500 J is not 105KJ it is 10,5KJ ! 105KJ = 105000J

 

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?

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.

 

Old thread, guys.

 

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.

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!