To calculate the work done in delivering water against a head, we can use the formula for work done, which is given by:
\[ W = F \times d \]
where:
- \( W \) is the work done (in joules),
- \( F \) is the force (in newtons),
- \( d \) is the distance the force is applied over (in meters).
In this case, the force required is due to the weight of the water being lifted, which can be calculated using the formula:
\[ F = m \cdot g \]
where:
- \( m \) is the mass of the water (in kilograms),
- \( g \) is the acceleration due to gravity (approximately \( 9.81 , \text{m/s}^2 \)).
First, we need to convert the volume of water (300 liters) into mass. The density of water is approximately \( 1000 , \text{kg/m}^3 \). Therefore, the mass of 300 liters of water is:
\[ m = \text{Volume} \times \text{Density} = 300 , \text{l} \times \frac{1000 , \text{kg/m}^3}{1000 , \text{l/m}^3} = 300 , \text{kg} \]
Next, we calculate the force due to the weight of the water:
\[ F = m \cdot g = 300 , \text{kg} \cdot 9.81 , \text{m/s}^2 = 2943 , \text{N} \]
Now we can calculate the work done to lift the water a height of 30 meters:
\[ W = F \times d = 2943 , \text{N} \cdot 30 , \text{m} = 88290 , \text{J} \]
Thus, the work done to deliver \( 300 , \text{l} \) of water against a head of \( 30 , \text{m} \) is approximately:
\[ \boxed{88290 , \text{J}} \]