Converting watt into temperature (celcius)

Converting watt into temperature (celcius)

I am trying to understand relationship between temperature and watt.

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 I am not an electrical engineer-so if I am missing something basic -please advise me.

Looking forward to replies.

Oh, no... you consider the CHANGE in temperature due to heating the air. Then degrees Celsius equal Kelvins. You do not start heating ar from zero Kelvins but from the ambient temperature. It can be 300 K or 20 C, it does not matter. Heat Joules increase the existing temperature by 112K or C.

What is about the thermal resistance Rth?
P = ( T2 - T1 ) / Rth

Bulbs are cooled by convection, how have you stopped the air moving? Typically the outside of the glass envelope is 80 degrees C when the ambient is 20.
Frank

Before calculating anything, you need to define a setup: Air volume and solids to be heated up, how good is the system isolated from the outer enviroment, otherwise what's the thermal resistance.

Usually a thermal circuit will involve multiple objects of defined heat capacity and connecting thermal resistances.