Respuesta :
Given:
Energy production rate, q = 45 kJ/L
Density, ρ = 0.77 g/L
Volume, V = 46 L
In consistent units, the total energy produced is
[tex]Q=(V\, L)*(\rho \, \frac{g}{L})*(q \, \frac{kJ}{g} )\\Q=(46\,L)*(0.77\, \frac{g}{L})*(45\, \frac{kJ}{L}) [/tex]
This will yield the energy in kJ.
Because the units are consistent, it will yield the correct answer.
All units may be converted to SI units, it required.
1 g = 10⁻³ kg in SI units
1 L is an accepted unit of volume in SI units.
1 kJ = 10³ J in SI units.
Energy production rate, q = 45 kJ/L
Density, ρ = 0.77 g/L
Volume, V = 46 L
In consistent units, the total energy produced is
[tex]Q=(V\, L)*(\rho \, \frac{g}{L})*(q \, \frac{kJ}{g} )\\Q=(46\,L)*(0.77\, \frac{g}{L})*(45\, \frac{kJ}{L}) [/tex]
This will yield the energy in kJ.
Because the units are consistent, it will yield the correct answer.
All units may be converted to SI units, it required.
1 g = 10⁻³ kg in SI units
1 L is an accepted unit of volume in SI units.
1 kJ = 10³ J in SI units.
Answer: 1592307.7 kJ
Explanation:- [tex]Volume=\frac{Mass}{Density}[/tex]
[tex]{\text {Volume of gasoline present}}=\frac{1g}{0.77g/ml}=1.30ml=0.0013L[/tex]
Thus 0.0013 L of gasoline when burned produces = 45.0kJ
46 L of gasoline when burned produces =[tex]\frac{45.0kJ}{0.0013L}\times 46L=1592307.7kJ[/tex]
