# Kelvin Temperature - Celsius Temperature Change

Hello,

I'm going to post a curious question.

Today I found a case where I have no idea what to do.

I'm working with the Thermal Conductivity and it's W/(m ºC) or W/(m K) then I don't know how to change from a way to the another one because kelvin and Celsius degrees are not proportional.

All my time in my studies I usually worked with Celsius degrees. When I had Kelvin I had the constants or variables in the appropriate units.

Can you help me with this?

Thank you.

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Kurdt
Staff Emeritus
Gold Member
K = ºc + 273.15

Yes, I know

This is ok if you want to pass from 20ºC to Kelvin.

but if you want to pass 0,08 W/(m ºC) ??

I think is not as easy as it looks

cheers

Meir Achuz
Homework Helper
Gold Member
Yes, I know
I think is not as easy as it looks
It's easier. The conductivity depends on Delta T, so it doesn't matter whether you use
Kelvin or Celsius.

FredGarvin
It's easier. The conductivity depends on Delta T, so it doesn't matter whether you use
Kelvin or Celsius.
That's it. You just care about the delta. When going between °C to K or °F to °R, you don't have to worry about the units. The deltas are the same.

Kurdt
Staff Emeritus
Gold Member
Ahh silly me. It was the thermal conductivities themselves you were referring to.

Then when I see data about a material property it doesn't matter if I see the properties in ºC or K...?

Yes, I agree with this, but it's strange to me.

thank you very much

russ_watters
Mentor
You said in the OP that they aren't proportional. They are more than proportional - the intervals are identical.

I know this is an old post, and even though the original question was answered, I found the answer hard to accept as did the original questioner.

I was taught to check all my unit conversions as a way to catch errors.

With the answer provided on this forum, it appear to me the following is being implied:

Given: .024 W/mK for the thermal conductivity of Air

So if all my other units of measure are in C and I don't want to convert them to K, you're telling me that it's legitimate to use .024 W/mC???? automagically????

Yes, I understand the scales of K and C have the same unit intervals BUT they start at different points.

I'd say that 273.15 units of difference is NOT negligible when moving from one scale to another and definitely not interchangeable.

Could you please give me some mathematical proof supporting the claim that the ratio of .024W/mK is the same ratio expressed in degrees C ?

...and if .024W/mK does not equal .024W/mC, then please answer the original question of how to convert the thermal conductivity factor from being expressed in K to being expressed in C.

negk

HEY,
So delta or the temp. difference whether you're in Kelvin or degrees Celsius is always the same, if you have to temps in Kelvin and the difference is 25 degrees it's the same as if you have a difference of 25 degrees in Celsius its just after you find the difference you then need to convert your aswer to either KElvin or CElsius, whatever you ar elooking for and slove the given equation!

russ_watters
Mentor
I know this is an old post, and even though the original question was answered, I found the answer hard to accept as did the original questioner.

I was taught to check all my unit conversions as a way to catch errors.

With the answer provided on this forum, it appear to me the following is being implied:

Given: .024 W/mK for the thermal conductivity of Air

So if all my other units of measure are in C and I don't want to convert them to K, you're telling me that it's legitimate to use .024 W/mC???? automagically????
What you're really doing here is multipllying by 1C/1K - but that's easy enough to do in your head; you don't need to write it down.
Yes, I understand the scales of K and C have the same unit intervals BUT they start at different points.

I'd say that 273.15 units of difference is NOT negligible when moving from one scale to another and definitely not interchangeable.
That has nothing to do with the proportionality of the scale.
Could you please give me some mathematical proof supporting the claim that the ratio of .024W/mK is the same ratio expressed in degrees C ?
The math is in the second post. Note the lack of a proportionality constant in front of the "ºC". That means the proportionality constant is 0.

You could also apply the whole equation if you want. So for example if you want to prove 10C-5C = 5K, you could do this:

(10C*K/C+273K)-(5C*K/C+273)=5K
...and if .024W/mK does not equal .024W/mC, then please answer the original question of how to convert the thermal conductivity factor from being expressed in K to being expressed in C.
The original question was answered correctly.

These statements really helped the bulb go on (tho dim it might be):

What you're really doing here is multipllying by 1C/1K - but that's easy enough to do in your head; you don't need to write it down.
That has nothing to do with the proportionality of the scale. The math is in the second post. Note the lack of a proportionality constant in front of the "ºC". That means the proportionality constant is 0.

And this math "proof" is exactly what I asked for to help illuminate my reasoning error:

You could also apply the whole equation if you want. So for example if you want to prove 10C-5C = 5K, you could do this:

(10C*K/C+273K)-(5C*K/C+273)=5K
The original question was answered correctly.
So, to make sure I really understand: for any two units of measure for which there is no rise/run (ratio or proportion), though the y intercept is different, can be used interchangeably in a physical constant (as long as my variables use one or the other without mixing unit of measures).

....seems elementary AFTER you spelled it out for me -thanks!

...makes me wonder if there are two units of measure (not a combination of units of measure) that have a logarithmic relationship or a three dimensional relationship rather than a y = mb+x relation...

negk

russ_watters
Mentor
There is an error in my post: the proportionality constant is 1, not 0.....but you got it now. And you're welcome.