Buffalo immersed himself from time to time, in liquid air.

I know we have all wondered, how much liquid air does it take to immerse a human body?

Probably more than you think.

Cooling Buffalo meant transferring his body heat to the liquid air. We will start by computing how much heat had to be transferred. Then we can easily compute how much liquid air it boiled.

First, how far did Buffalo have to be cooled?

When liquid air was dumped on Buffalo's warm body, the cold liquid boiled and kept boiling
until his body temperature fell below the boiling point of nitrogen, which is 77*K* or
‑320*F*. (By that point, the oxygen had already stopped boiling.)

(If you are feeling stress, wondering whether the author knows that mammals cannot survive freezing, you can relax.)

Before Buffalo reached the boiling point of nitrogen, he passed through the freezing point of water. At that point the water in his body froze, which released quite a lot of heat, which boiled off more air.

Let's compute how much energy must be transferred to reduce his body to the freezing point of water.

Assume Buffalo’s initial body temperature was 310*K* (99*F*). To reduce that to the freezing point of
water (273*K*) required a drop of 37*K*.

How much water did Buffalo contain? By happy coincidence, Buffalo weighed exactly 100 *kg*, which
we will assume was entirely water. (His body was probably not *entirely* water, but biology was not my
best subject.)

How much heat must be transferred to freeze that much water? The specific heat of water is 4 *kJ*/*kg*.
That is, cooling 1 *kg* of water by 1*K* will release
4 *kJ* (4000 Joules) of energy. So cooling Buffalo’s 100 *kg* by 37*K* released
100 *kg* × 4*kJ/kgK* × 37*K* = 14,800 *kJ*,
all of which must be transferred to the liquid air.

At that point Buffalo froze, releasing more energy, called *heat of fusion*. For water, the
heat of fusion is 334 *kJ/kg*, so freezing Buffalo released
100 *kg* × 334 *kJ/kg*
= 33,400 *kJ*.

Once he froze, how much energy must be transferred to cool him the rest of the way?

The specific heat of water-ice is only 2 *kJ/kgK*,
so cooling Buffalo the rest of the way from 273*K* to 77*K*,
released 100 *kg* × (273K ‑ 77K) × 2 *kJ/kgK*
= 39,200 *kJ*.

Thus, cooling Buffalo to the temperature of boiling air released total energy of
14,800 *kJ* + 33,400 *kJ* + 39,200 *kJ*
= 87,400 *kJ*. All that energy had to be transferred to the (now boiling) liquid air.

But the boiling air also had to cool the ice chamber, an aluminum box 5*m* square and 4*m* high.
How much energy did that require?

First, how much aluminum had to be cooled? Each of the chamber's walls had a surface area of 5*m* × 4*m*
= 20*m*^{2},
while its floor and ceiling had areas of 5*m* × 5*m*
= 25*m*^{2}, for a total surface area of
(4 × 20*m*^{2}) + (2 × 25*m*^{2})
= 130*m*^{2}.

The ice chamber’s walls and floor were 1*mm* thick (on top of perfect insulation that can be ignored)
so the total volume of aluminum to be cooled was
130*m*^{2} × 0.001*m*
= 0.13*m*^{3}.

In terms of mass, Aluminum has a density of
2.7×10^{3} *kg/m ^{3}*. So the walls and floor massed
2.7×10

(The ice chamber also contained equipment, which we will ignore because, unlike a wall, the equipment had small surface area relative to its volume, so its heat leaked out slowly.)

How far did the aluminum have to be cooled?
Fortunately, the chamber was not as hot as Buffalo's body, so it did not have to be cooled as much.
Before freezing, Buffalo let the chamber cool to 233*K* (-40*C*). From there the liquid air cooled
the aluminum down to 77*K*, a drop of 156*K*.

How much energy must be transferred to cool 351*kg* of aluminum by 156*K*?

The specific heat of aluminum is 0.91*kJ/kgK*,
so cooling the walls and floor released energy of
351*kg* × 156*K* × 0.91*kJ/kgK*
= 49,828*kJ*.

Finally, in addition to its walls, floor and Buffalo's body, the chamber also contained gaseous air, which had to be cooled but not immediately liquefied.

How much air? The chamber’s volume was
100*m*^{3}=100,000 liters of air.

How much energy must be transferred from the (warm) gaseous air to the (cold) liquid air to match their temperatures? To answer that question it will be simpler to measure the gaseous air in units of moles.

At standard temperature and pressure (STP=298*K* and one atmosphere) one mole of air (or any other gas)
will occupy a volume of 23 liters.
So the chamber's atmosphere (100,000 liters) works out to 100,000/23 = 4348 moles of air.

(If you think I cheated because the chamber was colder than STP, then good catch. But it's okay because the chamber contained one atmosphere at room temp. As the chamber cooled, its air pressure dropped but the number of moles remained constant.)

The molar heat capacity of nitrogen is
29 *J*/(*mol K*),
and oxygen is almost the same, so cooling the chamber's gaseous air released
4348 *mol* × 29 *J*/(*mol K*) × 156*K*
= 20*kJ*.

Thus the total energy released by cooling Buffalo and his ice chamber was
87,400*kJ* (his body) + 49,828*kJ* (the chamber walls) + 20*kJ* (the atmosphere.)
= 137,248*kJ*. This is the amount of energy transferred to the boiling liquid air.

How much liquid air boiled off?

The air was 80 percent nitrogen and 20 percent oxygen. Boiling off 1*kg* of oxygen will absorb
200*kJ* of energy (this is oxygen’s *heat of vaporization*) and nitrogen is almost the same at
213*kJ*. So boiling off 1*kg* of liquid air will absorb
(0.2 × 200*kJ*) + (0.8 × 213*kJ*)
= 210*kJ*.

Thus, the 137,248*kJ* released by Buffalo and his ice chamber boiled off
137,248*kJ* / (210*kJ/kg*)
= 654 *kg* of air.

In terms of volume, liquid air has a specific gravity of
0.84, so the liquid volume of 654*kg* is 654/0.84 = 779 liters, 206 US gallons.

In other words, when Buffalo dumped liquid air on his body, the first 206 gallons boiled off.

When that much air boiled off, how much space did it occupy? This is important, because it pressurized the chamber.

Under normal conditions, the volume of a gas does not depend on the mass of its molecules, only their number,
which is conveniently measured in moles.
At standard temperature and pressure (298*K* and one atmosphere) each mole of gas will occupy 23 liters.

How many moles of air were in the 654*kg* that Buffalo boiled off? One mole of nitrogen molecules
weighs 28*g*,
and one mole of oxygen molecules weighs 32*g*. (Remember they form double-weight molecules, N_{2} and
O_{2}.) So one mole of air (80% nitrogen) weighs about 29*g*, or
0.029*kg*. Thus, 654*kg* of air contained 654/0.029 = 23,000 moles.

How much gas is that?
At room temperature (298*K*), 23,000 moles will expand to a total volume of 23,000 × 23
= 530,000 liters, assuming pressure of one atmosphere.

However, the ice chamber was much colder than room temperature. Charles’ Law tells us that, as a gas
cools, it shrinks in proportion to its temperature (in deg *K*.) So at a temperature of 77*K*, the
air had shrunk to a factor of 77*K* / 298*K* or about 25% of its room-temperature volume.
Thus the boiling air expanded to a total volume of
530,000*L* × 77*K* / 298*K* = 140,000 liters, or
140*m*^{3} (cubic meters) assuming pressure of one atmosphere.

But the ice chamber could not hold 140*m*^{3}. The chamber is 5*m* square and
4*m* high,
so its volume was 5*m* × 5*m* × 4*m* = 100*m*^{3}.
What happens when you stuff a 100*m*^{3} space with 140*m*^{3} of air?

Boyles’ Law tells us that compressing 140*m*^{3} of gas into 100*m*^{3} will raise its
pressure by 140*m*^{3}/100*m*^{3} = 1.4 atmospheres.

Recall the chamber started with 1/4 atmosphere (one atmosphere shrunk to ¼ volume by the cold.) So the ice chamber briefly reached a pressure of 1.4+0.25 = 1.65 atmospheres, until the air cooled enough to liquefy and drain through the scuppers into the cold pit.

Later, when Buffalo warmed he needed air to breathe, provided by evaporation of the liquid air left in his tray. How much liquid air did Buffalo need left in his tray?

We know pressurizing the chamber to one atmosphere required 4348 moles of air, and each mole massed 29*g*.
So the total mass of air was
4348 × 29*g* = 126*kg*. Liquid air has a specific gravity of
0.84, so the liquid volume of 126*kg* is 126/0.84 = 150 liters, or 40 US gallons.
This is the amount of liquid air Buffalo needed in his tray after he froze, so he would have something
to breathe when he thawed.

In total, the mass of liquid air was 654 *kg* to freeze Buff and the chamber, plus 126 *kg*
for him to breathe, for a total of 780 *kg*, which is 1,716 lbs, 655 liters, or 246 gallons.

So Buffalo's air barrel had to hold 250 gallons.

The same equations apply to Luci. She weighed only 40*kg*, or 40% of Buffalo’s mass, but her ice chamber
was the same. Working through the arithmetic, she boiled off 400 *kg* of liquid air to cool her body and the
chamber walls. Her chamber was the same size, so she also needed 126 *kg* of liquid air left in her tray, to provide one atmosphere for her to breathe when she warmed. So her barrel held 400+126 = 526*kg*,
which is 1160 lbs, 630 liters, or 166 gallons.