Recent content by Mineral

I see, that makes more sense. Thank you.

I think it might be because the problem assumes all heat released over the two hours stays in the air inside the room rather than transferring to other objects, air outside, or converting into other types of energy? Not completely sure.

I agree, but that was the accepted answer, not sure why they made it that way.

The hall is not sealed, so the pressure remains atmospheric. But yes, the volume and moles remained constant so I used C_v as you suggested, I misspoke earlier.

I would hope the params are correct since the professor set it up. However, it seems my mistake was in not using the molar heat capacity for the ideal gas at constant pressure, so the correct answer was ΔT = (2QT_0)/(5PV) = 48 ℃. Thanks for your help.

Ah ok, this approach worked. I didn't realize I should use the molar heat capacity for the ideal gas since it's at constant pressure. Thank you very much.

Yes, since each person gave off 70 watts, to calculate the heat given off over two hours I multiplied 70 by the number of people (1800) and multiplied that times 7,200 seconds, which gave me 907,200,000 J.

I figured it was fine to convert from K to C at the very end since the value of the change in temperature would be the same in either unit.

My approach to the problem was to try using Q = cmΔt by rearranging it to solve for Δt=Q/cm. Since the power per person is 70 W, there are 1800 people in the concert hall, and the problem asks for the temperature rise over two hours, I multiplied 70 W×1800 people × (3600 sec × 2) which gave me...