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see this plzz...smthng usefull

Himanshu

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Posted on
4 Aug 2007 02:02:17 IST

Posts: 3834

4 Aug 2007 02:02:17 IST

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494

see this plzz...smthng usefull

Chemistry ,

Hieee

about the temp. scales.....

see 1st of all u have to know the ice pt. , the steam point and the no. of divisions...fr a particular scale....

See now

For ^oC ice point = 0 , steam point = 100 and no. of divisions = 100

For K(kelvin) , ice point = 273.15 , steam pt =373.15 , no. of div.= 100

For ^oF = ice pt = 32 , steam pt = 212 , no. of div. = 180

Fr R , ice point = 0 , steam pt = 80 , no. of div. = 80

now.....

to calculate rel. b/w 2 temp. scales...jus follow the foll. rel.

(temp. on 1 scale-ice pt.)/no. of divisions = (temp. on other scale - ice pt of that scale) / no. of divisions

eg. lets find rel b/w ^oC and ^oF AND K and R

let temp. on ^oc scale is C and on ^oF scale is F , K on kelvin scale and R

C - 0 / 100 = K - 273 / 100 = F-32/180 = R-0/80

K = C + 273.15..............as we all know

C = (5/9) (F-32)

and yes frm this we conclude that -40^oC = - 40^oF

hope this helps and its really imp.....

cheers!!!!!!!!!!!!!!!!!!!!!!!!!!

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waterdemon

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Joined: 11 Jun 2007 16:39:13 IST

Posts: 1048

4 Aug 2007 08:25:44 IST

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Good , Never know when these can come in handy.
Cheers!!!!!!!!!!!!!!

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