2

Denote by $a$ and $b$ temperatures measured in $^\circ \rm C$. My aim is to find their difference in Kelvin ($\rm K$). I thought of this question for fun after noticing that I can approach this problem in two different ways, resulting in two different answers.

Recall that $\mathrm K = \mathrm C + 273$. The difference of $a$ and $b$ in $^\circ \rm C$ is $|a - b|$.

  1. Convert the difference, namely $|a-b|$, to $\rm K$. $$|a-b|_\rm C = (|a-b|+273)_\rm K.$$ $\therefore$ The difference in $\rm K$ is $|a-b|+273$.

  1. By converting $a^\circ \,\rm C$ and $b^\circ \,\rm C$ to $\rm K$, we obtain $(a+273)\, \rm K$ and $(b+273)\,\rm K$ respectively. Therefore the difference in $\rm K$ is $$|(a\require{cancel}{\cancel{+\,273}})-(b\cancel{+\,273})|=|a-b|.$$ $\therefore$ The difference in $\rm K$ is $|a-b|$, equal to the difference in $^\circ \rm C$.

The approaches in both these calculations make total sense to me, but clearly each one yields a different answer.

What is the correct answer (if there is one) and what is the correct approach to solving it? Is there a general case as regards $a$ and $b$, or would such an approach only apply to specific examples of fixed quantities (e.g. $a=30$ and $b=60$).

I don't have to convert from $^\circ \rm C$ to $\rm K$, but similarly, I could work in Fahrenheit ($\rm F$); however I am not sure what would happen if I let $a$ or $b$ equal to $-40$, given that $-40^\circ \, \rm C = -40^\circ \, \rm F$.

Qmechanic
  • 220,844
Mr Pie
  • 135
  • 9

4 Answers4

7

Temperature difference is not the same thing as temperature. If two objects are at the same temperature, then their temperatures differ by $0^\circ$C. Would you say, "aha! $0^\circ$C = $32^\circ F$, so the two objects differ in temperature by $32^\circ$F"?

The correct approach is, of course, the second one. If you want to find the difference between two temperatures in a particular temperature scale, then express the two temperatures in that scale and then subtract.

Albatross
  • 72,909
1

Here is a visual for what’s happening:

enter image description here Temperature can be represented on a 1D number ray. The transformations that can be done on a ray is scaling and shifting. And this is exactly what the different temperature scales are related to each other by. Scaling and shifting.

Now the conversion between celsius and kelvin is just shifting thus the temperature differences are preserved. This is what your second approach tells you.

Superfast Jellyfish
  • 9,511
1

The problem with your method is that temperature difference and temperature cannot be used in similar fashion as you do with meter, etc.

Anyway $1° \text C=1 \text K$ when these represent temperature difference. Now this might be be a bit confusing if the context isn't given as $1°\text C =274.15\text K$.

So to resolve such confusion my book (Halliday Resnick Walker) has suggested that we represent temperature difference in non-absolute temperature scales such as Celsius and Fahrenheit by $\text C°$ and $\text F°$.

Then $1\text C° = 1\text K$ and $1°\text C =274.15\text K$ hence solving the confusion. It would be great if it is taken as convention by SI.

0
  1. Convert the difference, namely $|a-b|$, to $\rm K$. $$|a-b|_\rm C = (|a-b|+273)_\rm K.$$

To convert a temperature (the relationship is correspondence): $$x^\circ\mathrm C ≅ (x+273.15)\mathrm{ K} ≅ \left(\frac95x+32\right)^\circ\mathrm F.$$

To convert a temperature interval/difference (the relationship is equality): $$x^\circ\mathrm C=x^\circ\mathrm{ K}=\frac95x^\circ\mathrm F.$$

So, the correction for Method 1: $$|a-b|_\rm C = |a-b|_\rm K.$$

  1. By converting $a^\circ \,\rm C$ and $b^\circ \,\rm C$ to $\rm K$, we obtain $(a+273)\, \rm K$ and $(b+273)\,\rm K$ respectively. Therefore the difference in $\rm K$ is $$|(a\require{cancel}{\cancel{+\,273}})-(b\cancel{+\,273})|=|a-b|.$$ $\therefore$

Method 2 correctly takes the difference of the corresponding Kelvin temperatures.

ryang
  • 909