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Thanks for the answer. I need to rephrase my question. I have a source of energy, creating heat at the rate of 6000BTU/hour in a room. What to know how long does it take the room temprature changes from 86F to 96F

New Answer:

You have the heat input rate (6000 BTU/hr) and you know the size of the room. I don't know this last fact, so the answer will be "plug in your values". Also I assume heat can be distributed very quickly

Volume of Room: HxWxL

Volume of air in room = Volume of room - Volume of furniture (assume 800 ft3)

Hourly heating rate in ideal conditions =

0.02 BTU/hr x 1 ft3 = 1o F/hr

0.02 BTU/hr x 800 ft3 = (1/800)o F/hr

1 BTU/hr x 800 ft3 = (50/800)o F/hr = 0.062o F/hr

6000 BTU/hr x 800 ft3 = (6000 x 0.062)o F/hr = 372o F/hr

Assume heating rate is proportional to how long the heater is on (actually a bit off because air is cooled by walls and furniture etc.)

For a 800 ft3 room the heater would heat the room 10o F in (10/372) hrs, about (1.5 minutes. For larger or smaller rooms the time would be

(1.5 minutes) x 800/(Actual room volume) = Time to raise temperature by 10o F

--------------- Rate of temperature change is dependent on how rapidly the energy is supplies. Consder to 1 ft3 boxes of air, insulated so that no heat can enter or leave.

In the first box we introduce a small amount of gaseous explosive that will release 0.02 BTU of heat when ignited. In the second a finely divided iron powder that floats in the air and slowly rusts releasing the same amount of energy. The explosion and rusting are allowed to happen. Both boxes now contain air 1o F warmer than the initial condition. One box acheived the change almost instantaneously, the other over the course of hours or days.

In real life situations the heat is transferred through the box by convection and conduction. The rate depends on the temperature of the heater and the amount of agitation of the air.

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