Temperature Change in room

Xandu

I've been trying to solve this for about a week and I'm not sure where I'm going wrong.

The problem is to figure out how much the temperature of the auditorium increases when its full of people.

The room is 4350 cubic meters.
The meeting lasts 30 minutes (1800 seconds)
The specific heat capacity of air is 1.005 J/gram, or 1005 kJ/Kg
The mass of air is 1.2 kg/m
There are 823 people.
Each person gives off about 100 watts per second.

I've been using the equation Q = mc (delta T)

To get Q, I do 823 * 100 * 1800, which is 148,140,000

So then I do 148,140,000 / 1.2kg/m (1005kJ/Kg) ( 4350), which is 148,140,000 / 5,246,100 = 28.238 degrees Celsius, which is way off. them.

dav2008

Quick note: watts is a measurement of energy per time so I think you mean each person gives off 100 watts (100 joules per second). Also I think you mean the density of air is 1.2 kg/m³

Make sure you are using all standard units. (Notice that the specific heat is given as kiloJoules per kilogram.

Xandu

Yes, i meant 100 joules per second, my mistake, I'll try it again with 1.3 kg/m3. Thanks

dav2008

No sorry that was a typo on my part. If it's given as 1.2 then use 1.2.

My main point was to make sure your values are in joules.

Xandu

Oh, that makes sense, thanks.