Why do we not need to convert from degrees Celsius to Kelvin?

[AN630078]

Homework Statement

Hello, I was thinking, when performing calculations for specific heat capacity why do we not need to convert from degrees Celsius to Kelvin?

Relevant Equations

E=mc∆θ (I think that this may be applicable, ignore if you do not think it is relevant to my question)

I understand that the Kelvin scale begins at absolute zero and that the Kelvin scale may even be referred to as the absolute scale. As the units begin from a definite zero and increase with equal amounts of energy being added to the particles of the substance being measured (at equal intervals to those on the Celsius scale) yet the various temperatures are not compared with something else (like the freezing point of water which is defined as zero degrees Celsius). Would this apply to why we not need to change from degrees Celsius to Kelvin? Or am I thinking along the wrong lines and it is simply a matter of using SI. units etc.?

 

[Merlin3189]

You seem to understand that for heat capacity, it is the change of temperature that is significant. Since degrees Celsius and Kelvin are the same size, it doesn't matter which you use to measure a change in temperature.

A change, on any scale, is determined by taking one measurement from another. All that matters is the number of steps between them.

Say you had a water tank, you could measure the level starting from the bottom, or from the top, or from any arbitrary level. Each would (or could) give you a different number, eg. 7m above the bottom, 3m below the top or 2m above the halfway level. (h= +7, d= -3, or v= +2)
When the level changed, these might change to h= +4, d = -6 and v= -1,
so the change or difference would be; in h (+4) - (+7) = (-3), in d (-6) - (-3) =(-3), in v (-1) - (+2) = (-3)
The result is the same irrespective of where you start your scale, because the size of the units is the same.

 

[Gordianus]

Say you have water at 20 Celsius and you heat it up to 80 Celsius. What's the temperature increment?
Can you do the same case with Kelvins?

 

[Frabjous]

As long as your units are self consistent you can choose for your convenience. They are just a measuring stick.

 

[phinds]

Since degrees Celsius and Kelvin are the same size, it doesn't matter which you use to measure a change in temperature.

As long as your units are self consistent you can choose for your convenience. They are just a measuring stick.

Those statements are true BUT are widely misunderstood by journalists, even sometimes by folks designated as "science writers" because they will say, for example, that 80 degrees C is "twice as hot" as 40 degrees C. If they were using Kelvin, they would see that this is not true.

 

[Frabjous]

Those statements are true BUT are widely misunderstood by journalists, even sometimes by folks designated as "science writers" because they will say, for example, that 80 degrees C is "twice as hot" as 40 degrees C. If they were using Kelvin, they would see that this is not true.

Given that hot is measure of human perception, that could actually be true.

 

[etotheipi]

The short answer is that, if you denote the temperature in Kelvin by $\theta$ and in Celsius by $\theta'$, then the two only differ by a constant translation $\theta = \theta' + a$, so $\Delta \theta' = \theta_2 ' - \theta_1' = (\theta_2 + a) - (\theta_1 + a) = \Delta \theta$, in other words temperature differences coincide.

The longer answer is that Kelvin ($\mathrm{K}$) is a unit of thermodynamic temperature, defined by $1/T = dS/dE$, whilst Celsius ($\mathrm{C}^{\mathrm{o}}$) is not. Thermodynamic temperature is derived by imposing that the efficienty of a Carnot cycle depends only on the temperatures $\theta_H$ and $\theta_C$ of each reservoir, $e = f(\theta_H, \theta_C)$. Consider two heat engines operating between three reservoirs at $\theta_1 > \theta_2 > \theta_3$, then you must have$$e_{\text{overall}} = f(\theta_1, \theta_3) = f(\theta_1, \theta_2) f(\theta_2, \theta_3)$$from which you deduce the form of $f$ must be $f(a,b) = h(a)/h(b)$ for some function $h$, which we are free to choose so long as it is monotonic. The standard choice for $h$ is the identity map which, along with the reference state of $0 \mathrm{K} \leftrightarrow$ absolute zero, leads to the Kelvin scale.

To give an example, if you use an expression like $U = \frac{d_f}{2} \mathrm{k}_B T$, then $T$ is a thermodynamic temperature and you must use Kelvins!

 

[Frabjous]

The Rankine temperature scale is also a thermodynamic temperature scale

I believe that there is an implied relative to "absolute zero" in $U = \frac{d_f}{2} \mathrm{k}_B T$ and like equations that could be made explicit allowing additional temperature scales. I am not saying that adding in absolute zero makes a “convenient” choice. Also, adding the relative term is effectively defining a thermodynamic temperature. Am I missing anything else?

 

[AN630078]

You seem to understand that for heat capacity, it is the change of temperature that is significant. Since degrees Celsius and Kelvin are the same size, it doesn't matter which you use to measure a change in temperature.

A change, on any scale, is determined by taking one measurement from another. All that matters is the number of steps between them.

Say you had a water tank, you could measure the level starting from the bottom, or from the top, or from any arbitrary level. Each would (or could) give you a different number, eg. 7m above the bottom, 3m below the top or 2m above the halfway level. (h= +7, d= -3, or v= +2)
When the level changed, these might change to h= +4, d = -6 and v= -1,
so the change or difference would be; in h (+4) - (+7) = (-3), in d (-6) - (-3) =(-3), in v (-1) - (+2) = (-3)
The result is the same irrespective of where you start your scale, because the size of the units is the same.

Thank you very much for your informative response and explanation. I see, so it is acceptable to use either unit because the intervals of the scales are equal.

 

[etotheipi]

I believe that there is an implied relative to "absolute zero" in $U = \frac{d_f}{2} \mathrm{k}_B T$ and like equations that could be made explicit allowing additional temperature scales. I am not saying that adding in absolute zero makes a “convenient” choice. Also, adding the relative term is effectively defining a thermodynamic temperature. Am I missing anything else?

What does this mean, what is a 'relative term'? I can't parse it as it stands. Could you write your idea as an equation?

 

[Frabjous]

I am saying that instead of using a temperature scale with absolute zero equal to zero, we could write our equations relative to absolute zero without regard to the numerical value

$T\rightarrow T-T_{absolute zero}$

so

$U = \frac{d_f}{2} \mathrm{k}_B T \rightarrow \frac{d_f}{2} \mathrm{k}_B (T-T_{absolute zero})$

The point I am trying to make is that the requirement of absolute zero as a reference temperature does not limit our choice of temperature scales. I am not saying that carrying around $T_{absolute zero}$ is the most convenient thing to do.

 

[etotheipi]

Okay sure! But that is essentially equivalent to converting into Kelvin.

 

[Frabjous]

you must use Kelvins!

I probably over interpreted your comment.

Okay sure! But that is essentially equivalent to converting into Kelvin.

Yes it is, but my point is that there is nothing special about the Kelvin scale.

 

[etotheipi]

Yes it is, but my point is that there is nothing special about the Kelvin scale.

I think we have to disagree here. The thermodynamic temperature $\theta$ is defined such that the Carnot efficiency is$$e = 1- \frac{\theta_2}{\theta_1} = 1 - \frac{q_C}{q_H}$$i.e. of identical form to the equation involving the heats. Alternatively, you can define it by $\theta:= pV/N\mathrm{k}_B$. You can prove these are equivalent (I will do so if prompted...!). The second law of thermodynamics induces a canonical choice of temperature scale.

Of course you can always replace $\theta$ with $\theta_{\text{Celsius}} + 273.15$ if you want, but it's sort of obfuscating things.

 

[Frabjous]

I think we have to disagree here.

One more try.

Doesn’t the Rankine scale meet these criteria in a non-obfuscating manner?
That multiple temperature scales meet these criteria means that the Kelvin scale is not special.

Let’s create a new temperature scale, the PF.
Using your non-obfuscating postulate, let’s set absolute zero to 0PF.
From years of observing physicists, let’s set the freezing point of alcohol to 100 PF.
We now have a scale that is just as easy to use as Kelvin and that has a clear warning that bad things happen at 100PF. Given this additional feature, it is clearly superior to Kelvin.

 

[etotheipi]

The point is that you have for the function $f$ I defined in #7 that $1- f(\theta_1, \theta_2) = h(\theta_1)/h(\theta_2)$. There is not a unique choice for $h$, indeed you could set $h(x) = \sqrt{x}$, $h(x) = x$, $h(x) = \alpha x$, etc. In other words you can dilate the temperature scale by a constant factor and still have a valid thermodynamic temperature scale, sure. In that sense the "PF" is fine, as are "Rankines", etc., but these are all still thermodynamic temperatures as opposed to Celsius or Fahrenheit, which are not.

Of course you will also need to modify the units of constants like $\mathrm{k}_B$!

 

[hutchphd]

Of course you will also need to modify the units of constants like kB!

As a staunch defender of the Fahrenheit scale as being much more people-friendly, I propose that we switch to Rankine as the absolute scale of choice. Yes we will need to multiply $k_B$ by $5/9$ . Small price to pay. Modestly I propose to call this the "hutch S I" system.

Vive Fahrenheit !

 

[etotheipi]

The bit about "user-friendliness" reminded me of this:

 

[hutchphd]

As official F° spokesperson it is incumbent upon me to point out : you have interchanged the appropriate cold labels for F and C . 0° is "really cold out there" for F and "meh..." for C.

Fahrenheit Forever !

 

[Frabjous]

As official F° spokesperson it is incumbent upon me to point out : you have interchanged the appropriate cold labels for F and C . 0° is "really cold out there" for F and "meh..." for C.

Fahrenheit Forever !

It’s correct. Fahrenheit followers are just tougher.

 

[etotheipi]

As official F° spokesperson it is incumbent upon me to point out : you have interchanged the appropriate cold labels for F and C . 0° is "really cold out there" for F and "meh..." for C.

Ha! I didn't even realize that. Good catch