Converting Fahrenheit to CelsiusDate: 9/29/95 at 8:47:9 From: Anonymous Subject: Temperature Is there a temperature that is the same on both the Celsius and Fahrenheit scales? Date: 9/29/95 at 11:20:0 From: Doctor Ken Subject: Re: Temperature Hello! Nice question. There's a formula that you can use to get from Fahrenheit to Celsius: it's c = (f-32) * 5/9 . The way we get this formula is to say that when f is 32, c is 0, so we must have c = (f-32)k, where k is some constant. To find the constant, we know that when f = 212 (boiling), c = 100. So 100 = (212-32)k 100 = 180k k = 5/9 So with the formula c = (f-32) * 5/9, we can figure out when f = c. If you want more hints, don't hesitate to write back. -Doctor Ken, The Geometry Forum Date: 9/29/95 at 17:44:22 From: Doug Zink Subject: Re: temperature Dear sir, my question was if there is a temperature that is the same for both Celsius and Fahrenheit? Doug Date: 9/29/95 at 23:25:43 From: Doctor Ken Subject: Re: temperature Hello! The formula I gave you was one that converted Fahrenheit to Celsius. To solve your problem, you'd say "when is f equal to c," i.e. solve the equation x = (x-32) * 5/9. Does that make sense? -Doctor Ken, The Geometry Forum Date: 9/30/95 at 1:23:37 From: Doctor Andrew Subject: Re: temperature Dr. Ken said: >So with the formula c = (f-32) * 5/9, we can figure out when f = c. If there are values of c and f, where f=c, that satisfy that equation, and that are reasonable temperatures (greater than absolute 0), then there is such a temperature. Finding that temperature could be tough if you don't know much algebra, but we'd be glad to give you some help. You could also use your calculator to test different temperatures, until you get pretty close to the right one (try a value for f and see if you get the same value as the result of the formula). There is a value that is greater than absolute 0 that will work with the formula. Can you figure out what it is? -Doctor Andrew, The Geometry Forum |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/