Answer :
To solve this problem it is necessary to apply the thermal expansion in liquids for this particular case, the volumetric expansion.
The formula that allows us to calculate the volumetric expansion of steel is given by
[tex]\Delta V = V_0 \gamma \Delta T[/tex]
Where
[tex]V_0 =[/tex]Initial Volume
[tex]\gamma[/tex]= Thermal coefficient of volumetric expansion
[tex]\Delta T[/tex] = Change in Temperature
For steel and gasoline we would then have to,
STEEL:
[tex]\Delta V_s = V_s \gamma \Delta T[/tex]
[tex]\Delta V_s = (19.6gal)(36*10^{-6}/\°C)(41.5\°C-10.2\°C)[/tex]
[tex]\Delta V_s = 0.02208gal[/tex]
GASOLINE:
[tex]\Delta V_g = V_s \gamma \Delta T[/tex]
[tex]\Delta V_g = (19.6gal)(950*10^{-6}/\°C)(41.5\°C-10.2\°C)[/tex]
[tex]\Delta V_g = 0.582806gal[/tex]
Therefore the amount of gasoline spill out is
[tex]\Delta V =\Delta V_g-\Delta V_s[/tex]
[tex]\Delta V = 0.582806-0.02208[/tex]
[tex]\Delta V = 0.560726gal[/tex]