The correct answer is $\boxed{\text{C}}$, 45.0 tonnes.
The hauling capacity of a locomotive is the maximum weight of the train that it can pull. It is determined by the locomotive’s power, the coefficient of rail-wheel friction, and the weight of the locomotive itself.
The power of a locomotive is the amount of force that it can exert on the train. It is measured in horsepower (hp). The coefficient of rail-wheel friction is a measure of how much friction there is between the wheels of the locomotive and the rails. It is a dimensionless number between 0 and 1. The weight of the locomotive is the force that the locomotive exerts on the rails.
The hauling capacity of a locomotive can be calculated using the following formula:
$H = \frac{P}{f} + W$
where:
- $H$ is the hauling capacity of the locomotive in tonnes
- $P$ is the power of the locomotive in hp
- $f$ is the coefficient of rail-wheel friction
- $W$ is the weight of the locomotive in tonnes
In this case, the power of the locomotive is 22.5 hp, the coefficient of rail-wheel friction is 0.25, and the weight of the locomotive is 22.5 tonnes. Substituting these values into the formula gives:
$H = \frac{22.5 \text{ hp}}{0.25} + 22.5 \text{ tonnes} = 45.0 \text{ tonnes}$
Therefore, the hauling capacity of the locomotive is 45.0 tonnes.
Option A is incorrect because it is the weight of the locomotive itself, not the hauling capacity. Option B is incorrect because it is the coefficient of rail-wheel friction, not the hauling capacity. Option D is incorrect because it is the power of the locomotive, not the hauling capacity.