Temperature at Altitude

Atmospheric Layers

from Wikipedia

I recently partook in some international travel, which is also not a small reason why there hasn’t been a post in a while.  Curious thing about the rest of the world: they tend to uniformly use a system of units that’s widely regarded as superior to the set we use in the U.S. of A, handed down to us by our imperial fore-fathers.

Despite my general affinity for the SI system of units, I do have a soft spot for the Fahrenheit degree, seeing as how Celsius degrees are not, strictly speaking, SI, and Fahrenheit gives about twice the resolution of the Kelvin, allowing us to complain about the weather more precisely.

During my international flight, they give you a little display option to see your altitude and outside temperature; because I didn’t want to turn off my showing of X-Men: First Class, I peeked over at a neighbor’s monitor and saw that when we had climbed to an altitude of 10,000 meters, the outside temperature read the same value for Fahrenheit AND Celsius!

When would we expect that to happen?  There’s actually a pretty straight-forward equation for converting Fahrenheit to Celsius (if I sound surprised, it’s because so many thermodynamic quantities are not so easy to quantify):

T_{ \textdegree F} =\frac{9}{5} T_{ \textdegree C}+32 \textdegree F
note, the 9/5 technically has units of \textdegree F per \textdegree C

Always do a couple quick sanity checks when you get a new equation like this, which describes something you’re somewhat familiar with (also to check that I didn’t make a mistake).  Let’s check the freezing and boiling points of water.  Freezing is at zero Celsius, times 9/5, plus 32 comes out to 32 degrees F.  Okay, so far so good.  What about boiling?  100 degrees C times 9/5 is 180 plus 32 is 212 Fahrenheit.  Right.  Okay, so when will the two temperatures be the same?

Let’s substitute T_{\textdegree F} for T_{\textdegree C}:

T_{ \textdegree F} =\frac{9}{5} T_{ \textdegree F}+32 \textdegree F

Solving for T_{\textdegree F}:

-\frac{4}{5} T_{ \textdegree F} = 32 \textdegree F
T_{ \textdegree F} = -40 \textdegree F= T_{ \textdegree C}

Sure enough, the monitor showed -40 for both temperature units.  Does this make sense, based on our altitude?  I wasn’t sure at the time, but as it turns out, 10 km is towards the top of the troposphere, near the tropopause.  I was hoping to be able to calculate temperature as a function of altitude very accurately for comparison, but it turns out that while the temperature decreases fairly linearly with altitude, the “environmental lapse rate” is a function of latitude and atmospheric composition.  But, by and large, -40 degrees (Celsius or Fahrenheit) seems to be perfectly reasonable.

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