Question

How long can these people remain in class if the body temperature is `37 ^@ C` and a person feels uncomfortable above this temperature?

Consider a classroom that is roughly `5 m xx 10 m xx 3 m`, initially at `20 ^@ C` and 1 atm. there are 50 people in the closed, insulated class losing energy to the room at the average rate of 150 watt per person. How long can they remain in class if the body temperature is `37 ^@ C` and a person feels uncomfortable above this temperature. Heat capacity of air `= 3/7 R`

1. 4.34 minutes
2. 5.73 minutes
3. 6.86 minutes
4. 7.79 minutes

Answer 1

I think there's a typo in the problem. The molar heat capacity of air should be `~~ bb(7/2R)`, not `3/7R`.

The reason for this is that at room temperature, as per the equipartition theorem, there are three translational degrees of freedom and two rotational degrees of freedom for `"N"_2` and `"O"_2`, which make up `99%` of the air. This gives

`barC_V ~~ bb(3)/2R + bb(2)/2R = 5/2R`

and we know that `barC_P = barC_V + R`, so `barC_P = 7/2R`.

And from this, we find that out of the answer choices,

1. `"4.34 minutes"`
2. `"5.73 minutes"`
3. `"6.86 minutes"`
4. `"7.79 minutes"`

it will take `ul"6.86 mins"` for these people to get uncomfortable.

---

Cool problem. Since the room is insulated, we will have conservation of energy:

`q_"people" + q_"air" = 0`

where:

• `q_"air" = n_"air"barC_PDeltaT_"air"` is the heat flow into the air and `q_"people"` is the average heat flow out from the people.
• `DeltaT_"air"` is the change in temperature of the air in `""^@ "C"` (or `"K"`).
• `n_"air"` is the mols of air.
• `barC_P = 7/2R` is the given molar heat capacity of air in `"J/mol"cdot"K"`.

Assuming the air is an ideal gas,

`PV = nRT`

`=> n_"air" = (PV)/(RT)`

The volume of the room is:

`V = "5 m" xx "10 m" xx "3 m" = "150 m"^3`

or:

`V = 150 cancel("m"^3) xx (("10 dm")/cancel("1 m"))^3`

`= "150000 dm"^3 = "150000 L"`

Hence, the mols of air in the closed, insulated room is:

`n_"air" = ("1 atm" cdot "150000 L")/("0.08206 L"cdot"atm/mol"cdot"K" cdot "293.15 K")`

`=` `"6235.5 mols"`

Since the room is closed, the number of `"mols"` is constant throughout the heat distribution process.

The heat threshold that must be reached to make these people uncomfortable is:

`q_"air" = "6235.5 mols air" cdot 7/2 cdot "8.314472 J/mol"cdot"K" cdot ("310.15 K" - "293.15 K")`

`= 3.08476 xx 10^6 "J"`

Since `"150 W/person"` is released, or `"150 J/s/person"`, and there are `50` people,

`|q_"people"| = |-"150 J"/("s"cdotcancel"person")| xx 50 cancel"people"`

`=` `"7500 J"` of energy is released per second.

So, in order to release a total of `3.08476 xx 10^6 "J"`, this much time would pass:

`color(blue)(t) = 3.08476 xx 10^6 cancel"J" xx cancel"1 s"/(7500 cancel"J") xx "1 min"/(60 cancel"s")`

`=` `color(blue)ul("6.86 mins")`