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The first one:
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Fahrenheit - Celcius Conversion Formula

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


The second one:
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Celcius
a temperature scale in which 0' represents the freezing point of water and 100' the boiling point.
To convert a Celcius reading to Fahrenheit degree, multiply by 9, divide by , and add 32.
(also called "centigrade")

Fahrenheit
a temperature scale in which 32' represents the freezing point of water and 212' the boiling point. To convert to Celcius, subtract 32 from the Fahrenheit reading, multiply by 5, and divide by 9.

to re-state it:

Celcius to Fahrenheit:
m ~ 9
d ~ 5
a ~ 32

Fahrenheit to Celcius
s ~ 32
m ~ 5
d ~ 9

Namaste

2007-03-30 03:18:24 · 7 answers · asked by Anonymous in Science & Mathematics Other - Science

7 answers

As others have already pointed out, you have simply stated the same thing twice.

I prefer
F = 1.8*C + 32
C = (F - 32)/1.8

To convert C to F there is a quick mental computation (multiplying or dividing by nines and fives is not quick in my book) through a restructuring of the formula. It requires more steps, but is easier, sort of like if you want to multiply X by 99, it's much easier to multiply X by 100 and then subtract X from your result. Two steps instead of one, but much easier to do mentally.

To convert C to F:
Double the C value
Subtract 10% of that
Add 32

The above method is exact. If all you need is a quick estimate then just double the C value and add 30. This approximation is usually close enough for temperatures close to room temperature. It is exact at 50 degrees F (10 degrees C).

2007-03-30 05:31:55 · answer #1 · answered by dogsafire 7 · 0 0

I wont recommend any of these. I would prefer the following

C/5 = (F - 32)/9

and the reason for that is it makes more sense because the same technique can be used for any temperature scale.

The most general formula goes like this ..
(C - L)/(U - L) = (F - L)/(U - L)

where L and U are the lower and upper limits respectively. L = 0 degree Celsius and U is 100 degree Celsius. You have to remember these values for each scale for example for Fahrenheit L = 32 and U = 212 i.e

(C - 0)/(100 - 0) = (F - 32)/(212 - 32) which is equal to
C/5 = (F - 32)/9

Now for example if you choose a new scale like Kelvin you can make the formula again ..

(C - 0)/(100 - 0) = (F - 32)/(212 - 32) = (K - 273)/(373 - 273) and you can do all interconversion whenever you want for any two scales. Too long answer but it's good if you understand the fundamentals behind any formula.

2007-03-30 06:35:11 · answer #2 · answered by pushker 3 · 0 0

C = (F – 32) x 5/9 ---------1st
F = C x 9/5 + 32 -----------2nd
Actually both the formulae are easy and are one and same
Depending which value you have to calculate you can use one of the formula.
If you want to calculate value of C you can use formula 1st and to calculate F you use 2nd.
If you find any difficulty in remembering two formulae, you remember one of the formula and substitute the given values and calculate the other value.
Still I find 2nd formula easy to remember.
F = C x 9/5 + 32
If direct value of F is to be calculated & value of C is given, use the formula as such.
If value of C is to calculated, Subtract 32 from given value of F ang multiply this value by 5/9.

2007-03-30 03:52:51 · answer #3 · answered by Pranil 7 · 0 0

Technically, they are the same.

The first one is quicker to do for those who already know math. The user just plugs in F or C to get the corresponding C or F.

The second one is "friendlier." It gives steps for someone who is unfamiliar with math. Furthermore, the second formula explains the freezing point and boiling point of water, which is a nice touch. It's not necessary for someone to know that in order to figure out what 98 F is, but it's handy.

I would prefer the first one, but I already know how to perform algebra. I would use the second one to teach a 3rd-grader how to do algebra.

2007-03-30 03:21:39 · answer #4 · answered by Rev Kev 5 · 0 0

I would prefer a simple formulae-
c/9=f-32/9
Put all the available values and leave the one which you want to find out.

2007-03-30 04:35:01 · answer #5 · answered by harsh_goyal28 2 · 0 0

Hi. How does relate to Kelvin? One standard, based on an absolute starting data point (0 deg K) would be best, in my opinion.

2007-03-30 03:22:37 · answer #6 · answered by Cirric 7 · 0 0

They're the exact same, one's just in words instead of equations.

2007-03-30 03:21:20 · answer #7 · answered by Tim 4 · 0 0

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