## Friday, February 18, 2011

### Reason #792 I Love Mathematics

Is 80 degrees twice as warm as 40 degrees?  Discuss...

Ok, so obviously I wouldn't be posting this if there weren't some subtlety to this question.  I will say that three days ago I was pretty confident in my initial answer to this question.

1. Well, you didn't specify C or F, so the 80 could be in one system and the 40 in another. In that case definitely not twice as warm.

2. It's twice as much warmer than 0 degrees.

3. @gwaddell: If they're both in the same system?

@Jacob: I'm not sure I understand what you mean. Say more. What's your "it's"?

4. only in Rankine or Kelvin, not in Celsius or Fahrenheit.

5. Sorry, that was kind of an odd sentence. I meant that 80 degrees is only twice as warm as 40 degrees, relative to 0 degrees.

Which is to say that temperature scales are somewhat arbitrary. If whatever freezes at 0 degrees happened to freeze at 10 degrees, then the scale would have been defined differently. And you would be asking "Is 90 degrees twice as warm as 50 degrees?" So clearly "twice as warm" is just a relative measurement.

Do I get a gold star?

6. What's wrong with just "no"?

For a long answer: Celsius or Fahrenheit are measured on an interval scale, if you want to make that statement make sense, then you need to use a ratio scale like Kelvin.

7. All great answers. Gold stars for everyone. Maybe this was obvious to some of you, but this was clearly a misconception I held three days ago when, without much thought, I would have confidently defended 80 F as twice as warm as 40 F.

8. Is 80 degrees twice as warm as 40 degrees? Discuss...

This is clearly a relativist trick. If you keep your right hand on a 120 degree hot plate and your left hand in a 40 degree refrigerator for one minute and then place both hands in a bowl of 80 degree water, your right hand will feel two thirds as hot while your left hand will feel 200 percent warmer. Therefore,

2/3 = 200%

Paul Hawking
Blog:
The Challenge of Teaching Math
Latest post:
http://challenge-of-teaching-math.blogspot.com/2011/02/libbys-little-tigers.html

9. Ok I see what you had in mind. But I went even further (and in the process noticed what you did too). Right off the bat I started with the question: What does "cold" mean? If it is the common meaning of our perception of how cold something is (or how cold we feel) then definitely the answer is NO (especially in the Kelvin scale!). What other meanings of cold can I think of? something more tied to a directly measurable physical property perhaps. In an analog thermometer, how much has the Hg expanded? Probably not. How much has it risen from a certain point? This depends on the thermometer and where you start measuring. Some other physical property? How closer you are in freezing.. hmm what does "closer" mean? How much kinetic/thermodynamic energy do the molecules of water have? For Celcius or F definitely not double. We have to go to the Kelvin scale and even then I am not sure that Kelvin and energy are directly proportional.
In short, you have to define "cold" :-)

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11. Temperature, in this sense, is the measure of the average kinetic energy of the pieces of a system. Given that it is a linear measure, then a zero must be defined. This can be the zero of Kelvin, Celsius, or Farenheit; it doesn't matter which you choose. In all three systems, the pieces of an 80deg object will have twice the average kinetic energy than those of a 40deg object, relative to the arbitrary 0deg reference point you picked.

So yes, 80deg is twice as warm as 40deg.

Of course, if you are going to talk about the temp of a quantum object, or about the temperature of a light source, things get more complicated.

12. Jeff, I'm with you on the definition of temperature. But if you start in Kelvin with absolute zero, then 80 degrees Fahrenheit is only a small fraction farther from zero than 40 degrees Fahrenheit, so the additional amount of kinetic energy is much less than double, if that makes sense. I guess you mean that if you chose zero Fahrenheit as an arbitrary third point, then the difference in kinetic energy between your arbitrary point and 40 would be half the difference in kinetic energy between your arbitrary third point and 80. But I don't see how that would work with ANY arbitrary zero - it seems like it would only work when the difference from 0-40 equals the difference from 40-80.

Overall I would say yes to this question for Kelvin and no for F and C. I don't know enough about other scales to comment. Curious about the commenter above who said it depends on whether its an interval or a ratio scale. Can you explain that? (Also, I thought Kelvin was just Celsius shifted a bit to make zero equal to zero kinetic energy?)

13. I think most of the people here agree that in order to do such a comparison, you have to use a ratio scale (Kelvin being one example).

Why don't we take it a step further (or back if you will). Division and multiplication is not allowed on a non-ratio scale.