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Yesterday the temperature outside was 0.5 °C. Today, the temperature is 30 °C. 30 is 5300% more than 0.5, but today is obviously not 5300% hotter than yesterday.

In Fahrenheit, the temperatures are 33 °F and 86 °F, respectively. 160% hotter sounds more reasonable, but this argument uses the same logic as Celsius, just on a different scale.

Converting these temperatures to Kelvin, we get 273.70 K and 303.15 K, respectively. Since Kelvin is an absolute scale of temperature, can we correctly say that today is 11% hotter than yesterday?

Since temperature is relative, can we also claim that there is 11% more heat today?

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As you have proved, it depends on your reference point. All are correct. – ChrisF May 14 '13 at 21:22
@ChrisF I am not sure I have proven anything. I have shown relative differences in temperature, but am not sure how this translates logically to heat. – dlras2 May 14 '13 at 21:25
OK, "prove" is too strong. What you need to find out is what amount of energy 0.5 °C/33 °F/273.70 K represents and compare that to 30 °C/86 °F/303.15 K – ChrisF May 14 '13 at 21:27
@ChrisF Is that increase in energy 11% then? – dlras2 May 14 '13 at 21:32
up vote 22 down vote accepted

Yes, it's 11% hotter today than yesterday. Of the three temperature scales you discussed, only the Kelvin scale allows meaningful ratios to be calculated. Dividing two temperatures expressed in Celsius or Fahrenheit is simply a mistake. There are numerous physical examples where it makes sense to multiply or divide by a Kelvin temp, e.g., the ideal gas law and the Stefan-Boltzmann law.

A similar example of the incorrect use of a scale of measurement would be if we were to call chocolate 1, vanilla 2, and strawberry 3, and then say that chocolate minus strawberry was -2. It's not a scale that is defined so as to make subtraction meaningful.

BTW, with a scale like celsius, division could result in the statement that today is infinitely more hot than yesterday, if the temperature yesterday was 0 C.

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How does this relate to the increase in energy from ambient heat? Is it directly proportional to Kelvin, ie 11%? – dlras2 May 14 '13 at 21:35
@DanRasmussen It is. Multiply $T$ by $k/2$, where $k$ is the Boltzmann constant. This is then the heat energy per degree of freedom. For instance, if you have $N$ noninteracting point particles moving in 3 dimensions (3 translational d.o.f. and no internal ones), then you have internal energy $U = (3/2)NkT$, which is just what you learn is the case for a monatomic ideal gas. – Chris White May 14 '13 at 22:20
Get your nice ans badge" +1-->10th – ABC May 15 '13 at 2:27
An addition with regards to taking ratios of (for example) Celsius: If it was -1C yesterday, and +1C today, that would result in it being "-200% hotter" than yesterday. – Philip C May 15 '13 at 7:18

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