The efficiency of the car is 18%. The weight is 33 120 N, the total resistance to movement is 0.05

The efficiency of the car is 18%. The weight is 33 120 N, the total resistance to movement is 0.05 of the weight. For how many kilometers. 40 liters of gasoline will be enough for a car if it moves evenly. The specific heat of combustion of gasoline is q = 4.6 * 10 ^ 7 J / kg, the density of gasoline is 700 kg / m ^ 3

Mass of gasoline with volume V with density r:
m = Vr;
The amount of heat released during the combustion of gasoline with a mass of m:
Q = qVr, where q is the specific heat of combustion.
Part of the heat will be spent on doing useful work A:
A / Q = 0.18;
A = 0.18 * Q = 0.18 * qVr.
Vehicle resistance force:
F = 0.05 * P, where P is the weight of the vehicle.
Useful work A is equal to the cost of overcoming the resistance force:
A = Fs = 0.05 * Ps,
where s is the path length.
0.18 * qVr = 0.05Ps;
s = 0.18 * qVr / 0.05 * P = (0.18 * 4.6 * 10⁷ J / kg * 0.04 m³ * 700 kg / m³) / (0.05 * 33120 N) = 140,000 m
Answer: 140,000 m (140 km).