A 100W bulb left on for a day will raise the temperature of a 6X8X9' closet how many degrees?

Question

And what would the formula be?

And so would it make much of a difference to switch to a 14W flourescent for example?

And what would the formula be?

And so would it make much of a difference to switch to a 14W flourescent for example?

(I'd actually appreciate a serious answer. You can save those cutsie comments for someone else thanks)

So here's what I get following Steve H and Ra.Ge:

If
- air density is assumed 1.19 g/L;
- room contains 12,323 Litres (432 ft**3 = 12,323,000 cm**3)
- specific heat capacity of air is 1 joule per gram per deg K
- the 100W bulb generates 8,840,000 joules in a day

Then the 100 bulb (in a perfect closed system, I guess) would raise the temp by 8640000 / 14557 = 593 degrees ???

And with volume_watcher, if I assume an R-value of .32 for 3/8" drywall all around, I get:
100 X (6 X .32) = 192 degrees

What am I doing wrong??

Answer 1

Actually, there is an easy way to approximate your answer. A closet should reach steady-state equilibrium before the end of the day - that is, the heat added by the bulb would equal the heat lost by conduction through the ceiling, walls, etc. Once steady-state equilibrium is reached, it doesn't matter what the heat capacity of the air or the closet is - only the thermal conductivity of the walls, door, etc. and the surface area of each will be important.

Here is where effective "R-value" of the walls, ceiling, door, floor is useful. R-value is just the rate of heat transfer in units of K-m^2/Watt (or in British units ft^2 F/(BTU/hr)). The metric "R-value" is about 1/6 of the corresponding R-value in British units due to the conversion factors.

The formula for difference in heat at steady state (i.e., heat rise) would be approximately:

Delta-T = 100 Watts / (R1A1 + R2A2 + R3A3 + R4A4 + R5A5 + R6A6) where each of the 6 surfaces of the room are treated separately.

You need to know approximately what your insulation is for each surface - but a little research on the internet and comparison based on your closet's insulating properties should get you a halfway decent approximation.

Answer 2

The way to do this for real is to get a temperature sensor in the closet hooked to a computer and actually measure the temperature increase. There will be many complicating factors that will be hard to take into account theoretically.

But if you want to get an estimate of the maximum temperature increase, the way to do it is this --

First, evaulate the amount of energy in joules used up by the bulb in a day. That's easy. Just 100W times the number of seconds in a day.

Next, look up the heat capacity of air someplace. Somewhere you should be able to find data on the temperature rise of a certain mass of air when a certain amount of energy is added to it. With that in hand, next calculate the mass of air in your closet from the density of air and the closet dimensions.

Put it all together and compute the expected temperature rise of the closet. Of course, the real closet light bulb will heat not just the air, but the walls and any other contents, so the real temperature rise would be less.

Answer 3

If you were living in an Ideal world, all the energy dissipated by the bulb would be absorbed by the enclosed air and there would be no heat losses.

In which case, you need to find out how much energy is dissipated by the bulb which is given by the wattage multiplied by the number of seconds it is left ON for. This will give you the energy dissipated in Joules

now that you gave this much energy to the air, what is its temperature change? this is given by what is known as specific heat capacity. Specific heat capacity (SHC) is defined as how much would the temperature of a small volume of a substance would increase if you supplied one unit of heat energy to it. the heat units may eigher be in calories or joules. a calorie is 4.2 joules so you can use the appropriate conversion.

of if you have the SHC in "degreeC per joule per cubic centemeter", then you need to convert your volume of the closet to cubic centimeters. Then SHC * Joules * Volume in CC will give you the temperature increase in degree C, if you need it on some other temperature scale, you need to convert again.

all the complications lie in unit conversions, unfortunately there are too many units for the same thing. If you get the conversions right, the actual calculation is only a simple multiplication.

and now we step into the real world. all the energy dissipated by the bulb is not absorbed by the air. some of it is taken by the material enclosing the space. some of it is lost to radiation. and a lot of heat is lost everytime you open the space.

The only real way to find out is to acually setup the situation and measure using a thermometer, even then the results will not be uniform over several measurements. the losses also depend upon how cold is it ouside your enclosed space, something outside your control.

The scientifically correct but tedious method is to make several measurements over several sets of nominal conditions or either average them out or draw a probability distribution curve of the observations so that we can accurately predict what the next outcome might be.

edit:

Fluorescent lamps are much more efficient in producing light, ie for the same amount of light produced, produce much lesser heat. a 14 W fl lamp would take considerably longer to heat up the same space through the same temperature

Answer 4

not a scientist yet ill let u know when i get a badge

Answer 5

Growing alittle marihuana are we?