- #1
Mineral
- 3
- 0
- Homework Statement
- An audience of 1800 fills a concert hall of volume 2.2×10^4 m^3. If there were no ventilation, by how much would the temperature of the air rise over a period of 2.0 h due to the metabolism of the people (70 W/person)? Assume the room is initially at 293 K.
- Relevant Equations
- Δt=Q/(cm)
m=pV
E = Q (J) = power (W) * time (sec)
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 Q=907,200,000 J.
I used c=1006 J/kg * ℃ as the heat capacity for the air and 1.225 kg/m^3 for the density to find the mass m = 1.225 kg/m^3 × (2.2×10^4) m^3 = 26950 kg.
I plugged this information into the formula ∆t = Q/cm = (907,200,000 J) / (1006 J/kg * ℃ × 26950 kg) =33 ℃.
However, I’m aware this is wrong since I tried putting it into MyLab and it told me it was incorrect. I tried going over the problem several times, but I can’t find my error, the only thing I could think of is that maybe I used the wrong values for heat capacity or density of the air.
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 Q=907,200,000 J.
I used c=1006 J/kg * ℃ as the heat capacity for the air and 1.225 kg/m^3 for the density to find the mass m = 1.225 kg/m^3 × (2.2×10^4) m^3 = 26950 kg.
I plugged this information into the formula ∆t = Q/cm = (907,200,000 J) / (1006 J/kg * ℃ × 26950 kg) =33 ℃.
However, I’m aware this is wrong since I tried putting it into MyLab and it told me it was incorrect. I tried going over the problem several times, but I can’t find my error, the only thing I could think of is that maybe I used the wrong values for heat capacity or density of the air.