It shouldn't matter when I convert units in a calculation, the final answer should be the same. However:

$-30\ °C - 0\ °C = -30\ °C = (-30 + 273.15)\ K = 243.15\ K$

$-30\ °C - 0\ °C = (-30 + 273.15)\ K - (0 + 273.15)\ K = 243.15\ K - 273.15\ K = -30\ K$

where $T_C = T_K - 273.15\ K$, $C$ is Celsius, $K$ is Kelvin.

How come they're different?

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    $\begingroup$ You're not converting units, you're transforming scales here: Celsius and Kelvin degrees are the same, but the zeros of the two scales are different. So in the first attempt you calculate a temperature in Kelvin scale after applying a difference in Celsius degrees, the second is a calculation of a temperature difference in Kelvin degrees. Decide what you're calculating, then you'll know when to transform the scale. $\endgroup$ – CiaPan Jul 3 '18 at 14:00

A difference in temperatures isn't a temperature any more than a difference of dates is a date. You may convert temperatures to temperatures using your formula. Your formula does not convert differences in temperature to anything at all (in the same way that applying any other wrong conversion fails to produce a difference in temperatures).

  • $\begingroup$ Apparently some well-respected cookbooks have made statements like, "At altitudes above 5000 feet, increase cooking temperature by 10 degrees Celsius (40 degrees Fahrenheit)" $\endgroup$ – DJohnM Jul 3 '18 at 1:02
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    $\begingroup$ I think I disagree. Every temperature is "a difference in temperature". When we say the temperature is 30C, we are saying that is the difference between the current temperature and the freezing point of water. The key point is that we have an implicit base point that is different for the two scales. $\endgroup$ – Dave Costa Jul 3 '18 at 3:06
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    $\begingroup$ @DaveCosta : The base point vanishes in differences of temperatures. As a consequence, mentioning base points with differences of temperatures is misleading. The base point is necessary for interpreting temperature and is entirely absent interpreting difference of temperatures. The same happens for times vs. intervals, lengths vs. distance, and every other quantity that is measured relative to some arbitrary coordinate system. Conflating such pairs is incorrect. $\endgroup$ – Eric Towers Jul 3 '18 at 4:08
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    $\begingroup$ Because metres and centimetres have the same zero point, and the only difference is in scale. $\endgroup$ – ConMan Jul 3 '18 at 5:18
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    $\begingroup$ @A_for_Abacus : The displacement of $-30 \,\mathrm{m} = -3000 \,\mathrm{cm}$ from $0 \,\mathrm{m}$ appears coincident with the position $-30 \,\mathrm{m} = -3000 \,\mathrm{cm}$ because it is a displacement from $0 \,\mathrm{m}$. A displacement of $-30 \,\mathrm{m}$ from $10 \,\mathrm{m}$ has nothing to do with the position $-30 \,\mathrm{m}$, although it is exactly the same displacement. $\endgroup$ – Eric Towers Jul 3 '18 at 5:34

The second is correct, and the first is incorrect. You're looking at the difference of two measurements (and the difference in Kelvin will be the same as the difference in Celsius). In your first line, you're making a difference into an absolute measurement and then converting it, and that is incorrect.

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    $\begingroup$ Without knowing the original question, it is just possible that the first option could be right. For example, if you're told "The temperature is initially -30C and drops by 0C, what's the final temperature?", so that one temperature on the left hand side is relative. $\endgroup$ – origimbo Jul 2 '18 at 22:54
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    $\begingroup$ @origimbo: In fancier terms, the change is independent of origin but the points on the scale depend on the origin. So, yes, a complete question would be helpful. $\endgroup$ – Ted Shifrin Jul 2 '18 at 22:56

We have that

  • $0 °C\equiv273.15 \,K$

and therefore for $T_C$ expressed in $°C$

  • $T_K= T_C+273.15 $

therefore for $T_C=-30 °C$ we have

  • $T_K=-30+273.15=243.15\,K$

and then

$$-30\,°C-0\,°C=-30\,°C=243.15 \,K-273.15 \,K=-30\, K$$

  • 5
    $\begingroup$ You often write answers too hastily and don't pay attention to what the question actually is. $\endgroup$ – Ted Shifrin Jul 2 '18 at 21:59
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    $\begingroup$ Actually Kelvin isn't degree. It is just Kelvin! $\endgroup$ – Nosrati Jul 2 '18 at 21:59
  • $\begingroup$ @TedShifrin Thanks for your kind observation! You are right but sometimes I lost some important details! $\endgroup$ – gimusi Jul 2 '18 at 22:16
  • $\begingroup$ @user108128 Yes of course you are right! I've fixed that. $\endgroup$ – gimusi Jul 2 '18 at 22:16
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    $\begingroup$ Writing $-30 °C - 0 °C = -30 °C$ is misleading at best. I know you probably mean something among the line of $-30 °C - 0 °C = -30 °C_{\Delta}$, but you're better not expressing temperature differences in anything other than Kelvin. $\endgroup$ – Blackhole Jul 2 '18 at 22:40

The reason why it doesn't matter when you convert units is because typical unit conversions are multiplicative ratios. For example,

1 m = 1000 mm


1 m/1000 mm = 1

And we know that we can multiple any term in an equation by 1 without changing the meaning of the equation. But as you can see in your example,

1 C= 1 K + 273.15 

the unit conversion for temperature is a shift, not a factor change. Clearly the difference is that in one equation you are introducing 273.15 once and in the other twice. For this reason the only time when you can convert between degrees Celsius and Kelvins is when the variable represents a difference like delta C or dC.


The addition or subtraction of 273.15 is not actually a unit conversion; it is a scale conversion.

As units of heat, 1 degree Celsius = 1 Kelvin (and also, 9/5 degree Fahrenheit).

Addition or subtraction enters the picture because when we measure actual temperatures, we do it relative to a base point that is different for each scale. When we say it is 30C outside, we really mean it is 30 degrees Celsius above the freezing point of water, which is another way of saying it is 303.15 degrees Celsius above absolute zero.

So if you have a temperature measured in the standard Celsius scale -- e.g. "it's 30C outside today" -- you would add 273.15 to convert it to Kelvin. But if you have simply a quantity of degrees Celsius -- e.g. "the temperature increased by 30C overnight" -- the quantity is exactly the same in Kelvin.


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