# So im trying to find the Kelvin and Celcius.?

Well I think thats what its asking...Ok heres the 1st part of it:

I have the temp of bath water in Celcius: 1.1

and the Height of Air Column (relative vol.) (cm): 4.8

and I need to find the Temp(K)

& V/T (cm/K)

now how do I find them? Please explain! I have a test coming up and this might be on it!

AND I ALWAYS CHECK A BEST ANSWER..so someone will get 10 points! :)

If you know the celcius temperature, you just add 273.5 degrees to get the Kelvin temperature.

See the chart on the link for more conversions

Try here http://www.onlineconversion.com/...

You sound really confused.

First, K and Celsius are measure of temperature. There is definitely a section in your chem book on how to do conversions between different temp scales. You should check it out.

And I doubt if the bath water is 1.1 degrees C - that would be nearly freezing!

And the height of the air column is not going to affect the bath water. So what is the question?

You definitely need to spend some time with your teacher for extra help on this stuff.

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I have the temp of bath water in Celcius: 1.1

and the Height of Air Column (relative vol.) (cm): 4.8

and I need to find the Temp(K)

& V/T (cm/K)

now how do I find them? Please explain! I have a test coming up and this might be on it!

AND I ALWAYS CHECK A BEST ANSWER..so someone will get 10 points! :)

**Answer:**If you know the celcius temperature, you just add 273.5 degrees to get the Kelvin temperature.

See the chart on the link for more conversions

Try here http://www.onlineconversion.com/...

You sound really confused.

First, K and Celsius are measure of temperature. There is definitely a section in your chem book on how to do conversions between different temp scales. You should check it out.

And I doubt if the bath water is 1.1 degrees C - that would be nearly freezing!

And the height of the air column is not going to affect the bath water. So what is the question?

You definitely need to spend some time with your teacher for extra help on this stuff.

The answers post by the user, for information only, FunQA.com does not guarantee the right.

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