# 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 =

T

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 =

T

_{2}− T_{1}).Answer in units of K.Expert's answer

The conversion formula from Farenheight to Celsius is

[°F] = [°C] × 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× 9/5 + 32, T1 = -40 [°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 [°C]

Thus

T = T2 - T1 = 68.25- (-40) = 108.25 [°C].

[°F] = [°C] × 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× 9/5 + 32, T1 = -40 [°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 [°C]

Thus

T = T2 - T1 = 68.25- (-40) = 108.25 [°C].

Need a fast expert's response?

Submit orderand get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

## Comments

## Leave a comment