In Randall T. Knight’s textbook “Physics for Scientists and Engineers” in the first chapter on thermodynamics (Ch. 16: A Macroscopic Description of Matter) one of the first conceptual questions is stated like so:

The sample in an experiment is initially at $t_1$°C. If the sample’s temperature is doubled, what is the new temperature in °C?

The answer to that implies the reference to absolute temperature (for simplicity, to one significant figure, as the book required), i. e.:

$t_2°C = (T_1 × 2)K - 273 = (t_1°C + 273) × 2 - 273$.

I understand that the whole idea of this question to reiterate to the pupil that we ought to think of temperature as a specific physical property, and not in an everyday sense, but I am confused about the wording. In the section on temperature author writes:

We will begin with the commonsense idea that temperature is a measure of how “hot” or “cold” a system is. These are properties that we can judge without needing an elaborate theory. As we develop these ieas, we’ll find that temperature T is related to a system’s thermal energy. […] We’ll study temperature more carefully in Chapter 18 and replace these vague notions of hot and cold with a precise relationship between temperature and thermal energy.

That is, as yet we are still at the stage of “vague notions”. Few paragraphs later Knight introduces Kelvin scale but nowhere there or in between does he specifically insists on that idea I mentioned after the formula. At least not that I can see.

Am I missing something that should be obvious to me, and I fail to grasp it instantly because my brain is depleted by 50+ hours a week of studies? Am I just plain stupid, because to my eternal shame it took me way too long to figure it out? Or was there a better way to home onto that idea by an author: what do other textbooks say on that topic, how do they word this concept? Is the question logically flawed or confusing, or is it the common way of describing the situation in scientific literature?

  • 3
    $\begingroup$ Hi theUg, and welcome to Physics Stack Exchange! I changed your question title to make it more descriptive, although I wasn't really sure of a good phrasing. Feel free to change it again if you can think of something better. $\endgroup$
    – David Z
    Commented Dec 7, 2012 at 21:59

2 Answers 2


The question is meant to remind the student that the temperature relevant to physics is the absolute temperature in Kelvin, and not the temperature in relation to the triple point of water (in degrees Celsius).

That is, the zero point of absolute temperature is well-defined as the state without any thermal energy, and doubling absolute temperature is hence related to doubling the thermal energy. The zero point of the Celsius scale, however, does not relate to thermal energy in this way.

The formula given as a solution first converts degrees Celsius into Kelvin (absolute temperature), then doubles the temperature (something that is only sensible if done in Kelvin, not if done in degrees Celsius) and then converts it back to degrees Celsius.

That doubling the temperature is not a sensible thing to do in degrees Celsius, take the triple point of water minus $0.01^\circ\textrm{C}$, that is, $0^\circ\textrm{C}$. If you double this ‘temperature’, you get $0^\circ\textrm{C}$ again. However, if you double the thermal energy of the system and then convert it to degrees Celsius, you get $273.15^\circ\textrm{C}$ (modulo phase transitions occuring in between). If you do the same calculation in Kelvin, everything works out fine.

However, I have to agree that the question appears a little convoluted, as ‘doubling the temperature in degrees Celsius’ is not something I would even think of doing, and hence trying to lure the student into a trap by asking him to do so appears…strange.


The point of the exercise is simply that the answer is not $2t^{\circ} C$. They want you to understand that Celsius temperatures do not double because Celsius puts $0$ at the wrong reference point. It's an affine transformation, not a linear one. Imagine a plane, with units Kelvin on the $x$-axis and units Celsius on the $y$-axis. To convert from one to the other, you use a line that doesn't cross through the origin (because ${}^{\circ} C$ isn't $0$ when $K$ is 0). This line is of course roughly $C=K-273.15$. When you double temperature, you double your horizontal position (i.e, Kelvin). Seeing where that puts you on the line, it's clear that it didn't double your vertical position (i.e, Celsius).

If you understand why you have to convert to Kelvin first, then you've understood everything the exercise was trying to tell you.

  • $\begingroup$ See, the problem is while I (eventually) understood what I had to do, I am not quite sure the exercise combined with the text was trying (too hard) to tell me that. I think, explanation that these here answers present should be a core concept in the chapter, not the passing thought that the student gets nailed on when examined. $\endgroup$
    – theUg
    Commented Dec 7, 2012 at 21:46
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    $\begingroup$ By the way, kelvins are not degrees, they are just kelvins. ;) $\endgroup$
    – theUg
    Commented Dec 7, 2012 at 21:49
  • $\begingroup$ @theUg thanks for the catch on notation! I agree, and would have expected the text to mention this more explicitly since it trips students up quite often at first. I guess that's what phys.SE is for :) $\endgroup$ Commented Dec 7, 2012 at 21:55

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