What was the average force of water resistance to the movement of the steamer if it consumed 6.5 tons of coal

What was the average force of water resistance to the movement of the steamer if it consumed 6.5 tons of coal within 3 days at an average speed of 10 km / h? Motor efficiency 10%

Data: t (duration of the steamer movement) = 3 days. (in SI t = 259200 s); V (steamer speed) = 10 km / h (in SI V = 2.78 m / s); m (mass of coal consumed) = 6.5 t (in SI m = 6500 kg); η (motor efficiency) = 10% (0.1).

Constants: q (specific heat of combustion of coal) = 2.7 * 10 ^ 7 J / kg.

We express the average water resistance force from the formula: η = Ap / Q = Fc * S / (q * m) = Fc * V * t / (q * m), whence Fc = η * q * m / (V * t) …

Calculation: Fc = 0.1 * 2.7 * 10 ^ 7 * 6500 / (2.78 * 259200) = 24 375 N.