# The Kelvin-Celsius problem

Let's suppose we have temperatures 30°C and 35°C. Converting them to Kelvin we have 303.15K and 308.15K.

In the second case, the temperature difference is 5K. While in first case, temperature difference is 5°C, which is convertible to 278.15K.

This clearly is absurd. I have some ideas why it goes wrong, but could someone please provide a clear explanation? Cannot we use conversions on difference?

This seems to be a general problem with units which have additive relationship (Kelvin and Celsius) rather than a multiplicative one(metre -centimetres).

• You just unequivocally and clearly showed that no, you cannot use Celsius-Kelvin conversions on differences because the two scales differ by an offset. What exactly is the question about that? Commented May 1, 2016 at 10:32

Celsius and Kelvin are two scales that differs only for an additive factor, but the single increment corresponds to the same temperature difference. In other words, an object become "hotter" in the same way if you rise its temperature by 1K or 1°C.

You can use conversion formula in differences, just make sure you use it for both terms and keep in mind that in these cases you are referring to a temperature difference.

To make things more clear, consider the conversion to the Fahrenheit scale: 35°C - 30°C = 95°F - 86°F = 9°F which is telling you that the same temperature drop is represented by either 5°C or 9°F.

If that helps you, think about the potential gravitational energy on the Earth surface. That quantity is defined by choosing an arbitrary zero point, and this is exactly the case of Kelvin and Celsius scales. (Farenheit scale, instead, include a scale value).

• Wow, that actually works! What I am doing is choosing a reference point, finding a difference and then interpreting wrt another reference point. That is like calculating how much it would hurt falling one floor(11-10) and then thinking about the damage incurred falling 11 floors! Commented May 1, 2016 at 9:53

In an exam, Alice scored $50$, Bob scored $40$, Eve scored $10$. Now to raise the class mean, the teacher decided to add $20$ points to every students. So Alice's score becomes $70$, Bob's score becomes $60$, and Eve's score becomes $30$.

Now Alice complained to the teacher, "my score was higher than Bob's by $10$ marks, and you see, according to the conversion formula, as in Eve's case, $10$ marks should become $30$ marks. So my new score should be $30$ points higher than Bob's."

Edit:

Imaging more generally the teacher calculate the new score by

$$s'=f(s)$$

Then in general,

$$s'_1-s'_2=f(s_1)-f(s_2)\ne f(s_1 - s_2)$$

But it can be shown that equality holds if (and only if when imposing some other constraints, e.g., monotonicity, which should be true in your case)

$$f(s)=cs$$ for arbitrary $c$.

https://en.wikipedia.org/wiki/Cauchy%27s_functional_equation

Is that what you mean by "multiplicative"?

• Ah, lovely. But Alice thinks that the system is multiplicative but it isn't. Here the difference remains same, but in my question it does not. Commented May 1, 2016 at 9:49
• What do you mean by "multiplicative" - no, I think it has nothing to do with that; it's just nonsense, you get in an endless spiral of adjusting marks if you argument like that
– Ilja
Commented May 1, 2016 at 10:03
• Multiplicative like, a 10cm, whether it's difference or absolute is always 0.1 m. Commented May 1, 2016 at 13:48
• @Mathaholic yes, that's what it means, multiplication by constant. Commented May 1, 2016 at 13:50

temperature difference is 5°C, which is convertible to 278.15K.

This is where you go wrong.

A difference of 5°C is a difference. Express it as you will in Celcius, Kelvin, Rankine or even in Fahrenheit. But it is a difference.

Saying 5°C, which is convertible to 278.15K mean that you are not looking at a difference. Instead you are calculating how to show an actual temperature in another scale.

A temperature change of 1$^\circ$C is the same as a temperature change of 1K.

So if you start at 30$^\circ$C (= 303.15K ) and increase the temperature by 5$^\circ$C (5K) the new temperature is 30 + 5 = 35$^\circ$C (303.15 + 5 = 308.15K).

This seems to me like the general "Problem", that differences of Observables behave in a different way, than the observables itself do: One other example (that has nothing to do with additivity or multiplicativity, as you stated) are differences of frequencies: You know that you can convert a frequency to a wavelength by dividing c by the frequency:

\begin{align} \lambda = \frac{c}{\nu} \end{align}

Here c is the phase-velocity. However, if you want to calculate a difference in wavelength, you can't calculate it the same way: \begin{align} \Delta \nu = \nu_1 - \nu_2 \\ \Delta \lambda = \lambda_1 - \lambda_2 = \frac{c}{\nu_1} - \frac{c}{\nu_2} \neq \frac{c}{\lambda_1 - \lambda_2} = \frac{c}{\Delta \lambda} \end{align}

My interpretation of this Problem is that differences of observables are not the same as observables. They are given in the same units, like energy and work are given in the same units, but they do not physicaly represent the same thing: The Temperature of 30° C states that the body has a temperature of 30 °C because of the average energies of its particles. That is a property of nature, that we describe by a number. The 5 °C temperature-gap doesn't have such an aquivalent, so you can't convert it to the Kelvin units the same way you would do with a temperature.

The same goes for the frequencies: The frequency is a property of the wave, of nature. Because of the rules that govern nature, we know that its wavelength (also a property of the wave) can be calculated the way I stated it. A difference of frequencies itself doesn't describe the wave, so the rules that apply for waves can't be applied, and the difference of wavelengths can't be calculated the same way.

(Thanks to Drebin J.)

Let's suppose there's a building with many floors, labelled G, 1-10, and after that continues with A1, A2... A10, B1,B2 etc.

A1 is floor 1 but actually the 11th, while A2 is 2nd but actually 12th. (Similar to Celsius in Kelvin)

Supposing I jump from A2 to A1 (from the staircase), it would obviously hurt. The damage incurred corresponds to falling one floor (5 degree change!)

But instead as I am in the A-labeled floors, that one floor change must also be dealt with the same way others are i.e. By adding 10. So, I incur a damage of 11 floors!

The problem evident in this case, is that I change the frame of reference in this case, from A1 to ground.

It's only meaningful to describe the temperature of an object if the zeroth law of thermodynamics is true. There's probably no proof of the zeroth law of thermodynamics. See my answer at Is there any proof for the 2nd law of thermodynamics?. Assuming the zeroth law of thermodynamics, just because adding 2 distance given in meters gives the same result is converting them into kilometers then adding them the converting the result back to meters doesn't mean the same goes for the Celcius temperautre scale and the Kelvin temperature scale. For 2 objects of any temperature, English works in such a way that it's correct to say that a temperature is some number of degrees Celsius when the difference in the numbers that describe their temperature Celcius is that number. It's also correct to say their tempertuare difference is a certain number of Kelvins when the difference between the numbers that describe their Kelvin temperatures is that number. Although saying an object has a temperature of 5°C is another way of saying it has a temperature of 278.15 K, saying 2 objects have a temperature difference of 5°C is not another way of saying they have a temperature difference of 278.15 K. Because of the distributivity of multiplication over addition, given 2 distances in meters, you get the same answer if you just add them as if you convert them into kilometers then add them then convert the result back into meters.