First Law of Thermo: Audience body heat warming a concert hall

[Mineral]

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.

 

[DaveC426913]

Check units. You started with K and ended up with C. Or did you compensate for that and I missed it?

Is that 70W per person over 2 hours?

 

[Mineral]

Check units. You started with K and ended up with C. Or did you compensate for that and I missed it?

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.

 

[DaveC426913]

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.

Thats what I thought. Just checking.

What about the wattage? You multiplied 70 watts by 7200 seconds to get joules? OK. Just thinkin out loud here

 

[Mineral]

Thats what I thought. Just checking.

What about the wattage? You multiplied 70 watts by 7200 seconds?

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.

 

[DaveC426913]

OK, how do you know it's wrong? Obvs because you put it into MatLab, but how do you know you gave it the correct params for it to give you the correct answer?

(I mean, yeah. A sanity check looking at a 33C rise in temperature - that's a pretty good indication that your answer is indeed wrong, but still...)

 

[Chestermiller]

As an experienced thermodynamics practitioner, I would have approached this problem very differently.

From the 1st law of thermodynamics, we have $$\Delta U=Q-W$$Choosing the air within the auditorium as our system and assuming the volume of air is constant, we have $$\Delta U=Q$$where Q is the amount transferred from the people to the air in two hours.

If we assume that the gas is well-mixed and ideal, we have that : $$n=\frac{PV}{RT}=\frac{(100000)(2.2\times 10^4)}{(8.314)(293)}=99.3\times 10^6\ moles$$ $$\Delta U=nC_v\Delta T$$ and $$C_V=2.5R$$where n is the number of moles and C_V is the air heat capacity.

 

[Mineral]

As an experienced thermodynamics practitioner, I would have approached this problem very differently.

From the 1st law of thermodynamics, we have $$\Delta U=Q-W$$Choosing the air within the auditorium as our system and assuming the volume of air is constant, we have $$\Delta U=Q$$where Q is the amount transferred from the people to the air in two hours.

If we assume that the gas is well-mixed and ideal, we have that : $$n=\frac{PV}{RT}=\frac{(100000)(2.2\times 10^4)}{(8.314)(293)}=99.3\times 10^6\ moles$$ $$\Delta U=nC_v\Delta T$$ and $$C_V=2.5R$$where n is the number of moles and C_V is the air heat capacity.

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.

 

[Mineral]

OK, how do you know it's wrong? Obvs because you put it into MatLab, but how do you know you gave it the correct params for it to give you the correct answer?

(I mean, yeah. A sanity check looking at a 33C rise in temperature - that's a pretty good indication that your answer is indeed wrong, but still...)

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.

 

[DaveC426913]

I would hope the params are correct since the professor set it up.

Ah. When you said "I put it in to MyLab" I thought you meant you entered the whole thing and had it do your calc, not just that you entered your answer.

 

[DaveC426913]

so the correct answer was ΔT = (2QT_0)/(5PV) = 48 ℃.

That's even worse. That means the room reached an equilibrium temperature of 20+48=68C.

Not only would that kill everyone in the room in a matter of minutes but it seems physically impossible to accomplish from the body heat of people alone.

It seems to me, a ballpark estimate, with even the most aggressive values, suggests the room temp cannot reach, let alone exceed, 37C - average human body temp.

 

[Chestermiller]

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.

Who says the pressure is constant?? The volume and the number of moles are constant.

 

[Mineral]

Who says the pressure is constant?? The volume and the number of moles are constant.

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.

 

[Mineral]

That's even worse. That means the room reached an equilibrium temperature of 20+48=68C.

Not only would that kill everyone in the room in a matter of minutes but it seems physically impossible to accomplish from the body heat of people alone.

It seems to me, a ballpark estimate, with even the most aggressive values, suggests the room temp cannot reach, let alone exceed, 37C - average human body temp.

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

 

[DaveC426913]

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

But it has to be wrong. There has to be a flaw in the math. Any number of human bodys cannot raise the temp of any space to 68C.

 

[Mineral]

But it has to be wrong. There has to be a flaw in the math. Any number of human bodys cannot raise the temp of any space to 68C.

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.

 

[DaveC426913]

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.

Climb inside a barrel and close the lid (maybe add a breathing tube).

Same thing, smaller scale. It just cannot reach 68C, even in ideal thermodynamic conditions.

 

[Steve4Physics]

Human body temperature is, say, 40$^o$C. If the body is treated as a simple object which maintains its temperature at 40$^o$C, the room temperature can’t rise above 40$^o$C.

The question is at fault because it implies each body’s heat output is a constant 70W, independent of the room temperature. This is impossible. A body's heat output tends to zero as the temperature-difference between the body and its surroundings tends to zero.

 

[Chestermiller]

Human body temperature is, say, 40$^o$C. If the body is treated as a simple object which maintains its temperature at 40$^o$C, the room temperature can’t rise above 40$^o$C.

The question is at fault because it implies each body’s heat output is a constant 70W, independent of the room temperature. This is impossible. A body's heat output tends to zero as the temperature-difference between the body and its surroundings tends to zero.

This is definitely incorrect. The body gives off as much as necessary to maintain its temperature nearly constant, by means of chemical reactions which metabolize food.

 

[Chestermiller]

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.

The words " there were no ventilation" is meant to imply that the air volume and number of moles is constant, and the pressure increases (slightly) with increasing air temperature.

 

[Steve4Physics]

This is definitely incorrect. The body gives off as much as necessary to maintain its temperature nearly constant, by means of chemical reactions which metabolize food.

I'm a bit confused. In Post #18 I said "If the body is treated as a simple object which maintains its temperature at 40$^o$C". Isn't that entirely consistent with your (metabolic) description?

 

[Frabjous]

The mass of the people is greater than the mass of the air, so one should also account for the increase in temperature of the people which is beyond the scope of the problem as stated.

 

[erobz]

In order for the body to give off heat generated (metabolic) it must do so by some heat transfer process requiring $\Delta T$. If our bodies weren't able to sweat (adding water vapor to the surroundings) then in a sealed environment as the room (surroundings) increased temperature, so would we...and we die (at some point)?

You can experience this at Thanksgiving dinner in any home without AC (hopefully without the "and we die" part)!

 

[Mineral]

The words " there were no ventilation" is meant to imply that the air volume and number of moles is constant, and the pressure increases (slightly) with increasing air temperature.

I see, that makes more sense. Thank you.

 

[jtbell]

That's even worse. That means the room reached an equilibrium temperature of 20+48=68C.

Not only would that kill everyone in the room in a matter of minutes

Unless they’re Finns.

70C is actually on the low side for a traditional sauna!

 

[WWGD]

Unless they’re Finns.

70C is actually on the low side for a traditional sauna!

Huckleberry could really take the heat well!