Given :
If 2000 ft³ of air is crossing an evaporator coil and is cooled from 80°F to 60°F.
To Find :
The volume of air in ft³ exiting the evaporator coil.
Solution :
We know, relation between initial and final temperature and volume is given by :
[tex]\dfrac{V_1}{V_2}=\dfrac{T_1}{T_2}[/tex]
Here, temperature is in Kelvin .
We know, Kelvin temperature is given by :
[tex]K=(F-32)\times \dfrac{5}{9} + 273[/tex]
Putting F = 60° F,
[tex]K=(60-32)\times \dfrac{5}{9} + 273\\\\K = 288.56\ K[/tex]
Putting F = 80° F,
[tex]K=(80-32)\times \dfrac{5}{9} + 273\\\\K = 299.67\ K[/tex]
Putting all values, we get :
[tex]\dfrac{V_1}{2000}=\dfrac{60}{80}\\\\V_1=\dfrac{60\times 2000}{80}\ ft^3\\\\V_1=1500\ ft^3[/tex]
Hence, this is the required solution.