Comments on: An average question, just don't be mean.

Question: An average question, just don't be mean.

doozer_ex_machina — Fri, 15 May 2009 03:23:17 -0800

How much tap-temperature water (say 15 C) would you need to add to 1 litre of water to make a mass of water 80 C?

The instructions on my Aeropress say the best temperature say the best temperature for coffee is 80 C. I would imagine a simple average (1*100 + x*15 = 1 * 80 +x *80 => x~.3 litres ) of mass and temperature would not be correct considering I would get a different answer if I chose Kelvin, so what is the correct way? What is a good heuristic?

By: zippy

zippy — Fri, 15 May 2009 03:44:16 -0800

You have 1 liter of boiling water at 100 degrees C (373K).
You can add water that is 15 degrees C (288K).
You want the final amount of water to be 80 degrees C (353 K)

You're calculating a weighted average, where the liters are the weight, and the temperature is what's being averaged:

1 * 100C + x * 15C
----------------------- = 80C
1 + x

Solving,

100 + 15x / (1 + x) = 80

100 + 15x = 80 (1+x)
100 = 80 + 80x - 15x
20 = 65x
x = 20/65 liters

Works in Kelvin, too.

1 * 373K + 20/65 * 288K
----------------------------- = 353K = 80C
1 + 20/65

By: le morte de bea arthur

le morte de bea arthur — Fri, 15 May 2009 03:48:44 -0800

I got the same result as Zippy (on preview)

(1 * 100) + (x * 15) = (1 + x) * 80
which reduces to
20 = (x * 65)
or x = .3 litres

My calculation is the same in Kelvin:

(1 * 373) + (x * 288) = (1 + x) * 353
373 + 288x = 353 + 353x
(subtract 353 from both sides, then subtract 288x from both sides)
20 = (x * 65), i.e. the same result...

By: doozer_ex_machina

doozer_ex_machina — Fri, 15 May 2009 04:01:49 -0800

Thanks for the quick response. I feel a bit shamefaced I didn't go to the effort of checking it in Kelvin before stating it would be different :-(

By: RobinFiveWords

RobinFiveWords — Fri, 15 May 2009 16:01:30 -0800

The underlying assumption here is that the temperature scale is linear -- i.e., that the difference between 0 and 50 degrees is "equal" to the difference between 50 degrees and 100 degrees. I think this is true as long as the specific heat of water (the heat energy needed to raise the temperature of a certain mass of water by a certain amount) is relatively constant over the range we're considering.

According to this chart, it seems there's less than 1% deviation in specific heat capacity between the boiling point and whatever the baseline point is (either room temperature or the freezing point). So the scale is effectively linear, and the weighted average calculation is valid.