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# Answer to Question #1506 in Mechanics | Relativity for jon allen

Question #1506
Initially, an object has a temperature whichhas the same numerical value in degrees Celsius and degrees Fahrenheit. Then its temperature is changed so that the numerical value
of the new temperature in degrees Celsius is 5
times as small as that in Kelvin.
Find the change in the temperature (change in T =
T2 − T1).Answer in units of K.
Expert's answer
The conversion formula from Farenheight to Celsius is
[&deg;F] = [&deg;C] &times; 9/5 + 32
Denote the tempetrature in Celsius as T.
As the temperature has the same numerical value in degrees Celsius and degrees Fahrenheit
T1 = T1&times; 9/5 + 32, T1 = -40 [&deg;C].

Then its temperature is changed so that the numerical value of the new temperature in degrees
Celsius is 5 times as small as that in Kelvin:
T2 = (T2+273)/5, T2 = 68.25 [&deg;C]

Thus
T = T2 - T1 = 68.25- (-40) = 108.25 [&deg;C].

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