Answer: 0.4911 kg
Explanation:
We have the following data:
[tex]\rho_{0\°C}= 730 kg/m^{3}[/tex] is the density of gasoline at [tex]0\°C[/tex]
[tex]\beta=9.60(10)^{-4} \°C^{-1}[/tex] is the average coefficient of volume expansion
We need to find the extra kilograms of gasoline.
So, firstly we need to transform the volume of gasoline from gallons to [tex]m^{3}[/tex]:
[tex]V=8.50 gal \frac{0.00380 m^{3}}{1 gal}=0.0323 m^{3}[/tex] (1)
Knowing density is given by: [tex]\rho=\frac{m}{V}[/tex], we can find the mass [tex]m_{1}[/tex] of 8.50 gallons:
[tex]m_{1}=\rho_{0\°C}V[/tex]
[tex]m_{1}=(730 kg/m^{3})(0.0323 m^{3})=23.579 kg[/tex] (2)
Now, we have to calculate the factor [tex]f[/tex] by which the volume of gasoline is increased with the temperature, which is given by:
[tex]f=(1+\beta(T_{f}-T_{o}))[/tex] (3)
Where [tex]T_{o}=0\°C[/tex] is the initial temperature and [tex]T_{f}=21.7\°C[/tex] is the final temperature.
[tex]f=(1+9.60(10)^{-4} \°C^{-1}(21.7\°C-0\°C))[/tex] (4)
[tex]f=1.020832[/tex] (5)
With this, we can calculate the density of gasoline at [tex]21.7\°C[/tex]:
[tex]\rho_{21.7\°C}=730 kg/m^{3} f=(730 kg/m^{3})(1.020832)[/tex]
[tex]\rho_{21.7\°C}=745.207 kg/m^{3}[/tex] (6)
Now we can calculate the mass of gasoline at this temperature:
[tex]m_{2}=\rho_{21.7\°C}V[/tex] (7)
[tex]m_{2}=(745.207 kg/m^{3})(0.0323 m^{3})[/tex] (8)
[tex]m_{2}=24.070 kg[/tex] (9)
And finally calculate the mass difference [tex]\Delta m[/tex]:
[tex]\Delta m=m_{2}-m_{1}=24.070 kg-23.579 kg[/tex] (10)
[tex]\Delta m=0.4911 kg[/tex] (11) This is the extra mass of gasoline
