Breaking the rules of math. The temperature addition paradox!

This viral meme is an interesting math question. Obviously 0C = 32F. So then 0C + 0C = 64F, right? Clearly this is not correct, so where is the mistake? And what is the proper way to add temperatures?

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[M_Sp_]

Adding a temperature instead of a delta temperature to a temperature is like adding the 23rd of March to the 19th of December to get the 42nd of a new development month.

[UCXEOLxnaMJhtUsuNXhlmQ]

This isn't a paradox, just a fundamental misunderstanding of how relative systems work

[kburke]

You should add that the real reason you can’t do this is because the scales have different zero values. Degrees C and Degrees F are themselves not quantities but deltas above arbitrary zero points. You could add Kelvins and degrees Rankine in the way you suggest and it would work because both have the same zero point. For an extreme case add 16 degrees F too itself, then do the same with minus 10 C. What does it even mean to add negative temperature - are you adding heat content or specifying a change- that’s the ambiguity.

[GlizzyTrefoil]

To clear up why in thermodynamics we never just add temperatures, I recommend you to check out what intensive and extensive properties are. It is a straight forward answer, no need to be afraid of physics. Edit: "intrinsic" confused with "intensive"

[v_iancu]

That's why scientists measure temperature in Kelvins. It really measures heat.

[charcoalPanda]

I got shivers down the spine each time you say "ChA-A-A-Ange". Some serious teacher energy here.

[eddyk]

If you want to add temperatures, you've got to do it in Kelvin because 0 Kelvin is absolute zero and therefore has zero heat energy. 0C is just the freezing point of water, but still has 273C of heat energy

[Uejji]

In university, we learned about different kinds of values:
-Nominal: data can only be described
-Ordinal: data can be put in order
-Interval: distance between data can be measured (data can be meaningfully subtracted and these intervals can find new values)
-Ratio: data has a natural zero (data can be added, subtracted, multiplied and divided to each other)

Degrees C and F are interval values, so even though they have a value labeled "0, " it does not signify a zero value, so adding temperature values together is meaningless (a number being defined as its distance from zero value). Kelvin, however, *is* a ratio value, because it has a zero value at 0.

[wiggles]

I was hoping for some discussion about linear maps. A linear map is a map f from W to V such that f(x+y) = f(x) + f(y), and f(cx) = cf(x), for all vectors x, for all vectors y, and for all scalars c.

An example of a linear map is one that tells you how much tax gets added. For example, 10% tax means if something has a cost y in dollars, then f(y) = 1.1y. Checking the first condition:
f(x+y) = 1.1(x+y) = 1.1x + 1.1y = f(x) + f(y), thus the first condition is met. Checking the second condition:
f(cx) = 1.1cx = c1.1x = cf(x), so 2nd condition is met.
Thus, if you want to buy an apple at $5, and an orange at $6, you can find the tax on $11, or you can find the tax on $5, the tax on $6, and add the taxes.

For temperatures, let's see if we can do the same. Can we add two temperatures in Celsius and then convert to Fahrenheit, and get the same answer as converting to Fahrenheit before adding the temperatures? The map that goes from Celsius to Fahrenheit is g(T) = 1.8T + 32. Lets check if it's a linear map:
g(x+y) = 1.8(x+y) + 32 = 1.8x + 1.8y + 32, and
g(x) + g(y) = 1.8x + 32 + 1.8y + 32 = 1.8x + 1.8y + 64.
We can tell that g(x+y) != g(x) + g(y), so using the properties of a linear map is unjustified and will fail in almost every case.

Just to be clear, we let x = 0 degrees Celsius, and y = 0 degrees Celsius, and we tried to assert that g(x+y) = g(x) + g(y). Obviously we should expect that statement to be incorrect.

[ado-]

I've watched your videos since I had been preparing for highschool entry exams.You've definitely helped me in some way, teaching me to observe things from multiple perspectives, and I'm grateful for that.Keep up the good work and Mind Your Decisions :D

[CONNELL]

Adding temperatures is not physically meaningful. If my house is at 70 but my neighbor's is at 72, you cannot say that our combined temperature is 142. But you could legitimately say that our average temperature is 71, even though calculation of this quantity involves an addition

[WestExplainsBest]

4:07
I like how he specifies that you are adding a CHANGE, not an additional temperature reading.

[sebastiengross]

Not a math paradox but it's physics. Temperature is an intensive property: it does not depend on the amount of substance which was measured so you cannot add their values. Take one glass of water at 20ºC and mix it with an other glass at 20ºC you do not end up with water at 40ºC.

[dancoulson]

You need to convert to kelvin scale for addition of heat energy to work correctly.
0*C = 273*K
so 0*c + 0*C = 272*C (546*K)
Or another way of putting it - If you were to have a material at 0*c and double how much heat energy it contains, it would be 272*C

[Deggery-OneAboveAll]

I like how you took 68°F so you can convert it to 69°F 🤣

[Tiqerboy]

The Celsius and Farenheit scales measure relative, not absolute temperature. It's the same reason why 2­°C is not twice as warm as 1°C

[Remioxen]

That's less of a mathematical problem and more of a physics/thermodinamics problem

[capchemist]

Kelvin system: this is why you can't have nice things

[PhilWalton]

Solution: measure temperature in Kelvin.

[JackPullen-Paradox]

We could look at the equation with degrees Fahrenheit on the Y-axis. Then double the equation. The Centigrade and Fahrenheit scales would then agree with the arithmetic in the paradox. And I must ask what would happen if we took the mean of n Fahrenheit temperatures and compare the mean to the mean of the corresponding Centigrade temperatures? So, (0 + 0 + 0)/3 = 0 C= (32 + 32 + 32)/3 = 32 F, but (0 + delta-0 + delta-0)/3 = 0 C <> (32 + 0 + 0)/3 F. So, when considering the paradox, if one wanted to perform arithmetic on C and F, the equation approach would allow it. But the physical interpretation would require that we manipulate quantities as energy, and that means that the two temperature measures must actually be equivalent as far as any particular interpretation is concerned. The problem with the delta approach is that it necessarily ignores the zero offset of 32 degrees. Of course, for serious use, the delta approach is closer to the truth than the equation approach. But for recreation, it may be good to look at both.