First, it is necessary to calculate the energy consumed by the bulb during the weekend:
[tex]P = \frac{E}{t}\ \to\ E = P\cdot t = 75\ W\cdot 48\ h\cdot \frac{3\ 600\ s}{1\ h} = 1,296\cdot 10^7\ J[/tex]
(It is important to express time in seconds).
The energy associated with water is gravitational potential energy and we can write this type of energy depending on volume of water:
[tex]E = m\cdot g\cdot h\\ \rho = \frac{m}{V}\ \to\ m = \rho \cdot V\\ E = \rho \cdot V\cdot g\cdot h[/tex]
Clearing volume in the previous equation:
[tex]V = \frac{E}{\rho \cdot g\cdot h} = \frac{1,296\cdot 10^7\ J}{10^3\frac{kg}{m^3}\cdot 9,8\frac{m}{s^2}\cdot 40\ m} = \bf 33\ m^3[/tex]
