1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Help with thermodynamics!

  1. Dec 9, 2013 #1

    An audience of 1800 fills a concert hall of volume. If there were no ventilation, by how much would the temperature of the air rise over a period of 2.0h due to the metabolism of the people (70 W/person)? Assume the room is initially at 293 K.


    Q = m*C*deltaT (?)
    Convection equation (?)


    I'm pretty stumped on this question, I tried to use Q=m*C*deltaT (by calculating the mass of air the room) but I don't think that's the right way to do it. I'm pretty sure this is a convection problem but I can't apply the convection equation because there's some missing values.
  2. jcsd
  3. Dec 9, 2013 #2


    User Avatar
    Gold Member

    I think you are missing some information from the problem statement. You should be able to calculate the heat transferred to the air from the metabolism of the people as a first step.
  4. Dec 9, 2013 #3
    Wouldnt that be (70 W)(1800 people)(2 hrs)(3600 secs)= 9.072*10^8 J?
  5. Dec 9, 2013 #4


    User Avatar
    Gold Member

    Yes, now you can find the change in temperature of the air using Q = cmΔT and the criterion in the question. Was the volume of the hall given in the problem statement? It seems the first sentence was cut off short.
  6. Dec 9, 2013 #5
    Well if you have the volume, which you didn't write in the problem description, I see no problem in using Q=m*C*delta
  7. Dec 9, 2013 #6
    Oops I left that out when copying the question over, it should be 2.2*10^4 m^3 (2.2*10^7 L).

    So (2.2*10^7 L)/(22.4 mol/L) = 9.82*10^5 mol of air. Air has a molar mass of 28.97 g/mol so the total mass of the air in the room should be (9.82*10^5 mol)*(28.97 g/mol) = 2.85*10^7 g = 2.85*10^4 kg.

    Q = m*c*deltaT (Using 1006 J/kg*K for the specific heat of air)

    9.072*10^8 = (2.85*10^4)*(1006)*(deltaT)

    deltaT = 32 K

    This is not right though.
  8. Dec 9, 2013 #7


    User Avatar
    Gold Member

    What is the right answer? All your working is correct. Alternatively, if you were given the volume of the hall, then knowing the density of air ≈ 1.2 kg m-3 gives you the mass of air in the room. The result is of the same order of magnitude, as expected, and the change in temperature deviates by 2K.

    Did the values you use for the quantities given in the question or did you find them elsewhere?
  9. Dec 9, 2013 #8
    The only value that wasn't given was the specific heat of air (1006 J/kg*K), which I found online. The problem is from an online homework, so I don't know what the actual answer is, I just know that what I found isn't right.
  10. Dec 9, 2013 #9
    I found the problem in my textbook, and the answer is 48 C. I don't understand this answer, since I just did a recalculation with the correct values for specific heat and density of air at 293 K and still got the same answer as before.
    Last edited: Dec 9, 2013
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted