1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Temperature Change in room

  1. Apr 19, 2006 #1
    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.
  2. jcsd
  3. Apr 19, 2006 #2


    User Avatar
    Gold Member

    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.
    Last edited: Apr 19, 2006
  4. Apr 19, 2006 #3
    Yes, i meant 100 joules per second, my mistake, I'll try it again with 1.3 kg/m3. Thanks
  5. Apr 19, 2006 #4


    User Avatar
    Gold Member

    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.
  6. Apr 19, 2006 #5
    Oh, that makes sense, thanks.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Temperature Change in room