Question

Water falling from a 50-m high fall is to be used for generating electric energy. If $1·8\times {10}^5\mathrm{kg}$ of water falls per hour and half the gravitational potential energy can be converted into electrical energy, how many 100 W lamps can be lit with the generated energy?

Answer 1

Given:
$$\begin{gathered} \mathrm{Height},h=50\mathrm{m} \\ \mathrm{Mass}\mathrm{of}\mathrm{water}\mathrm{falling}\mathrm{per}\mathrm{hour},m=1.8\times {10}^5\mathrm{kg} \end{gathered}$$
Power of a lamp,
$$\begin{gathered} P=100\mathrm{watt} \\ \mathrm{Potential}\mathrm{energy}\mathrm{of}\mathrm{the}\mathrm{water}, \\ \mathrm{P}.\mathrm{E}.=mgh \\ =1.8\times {10}^5\times 9.8\times 50 \\ =882\times {10}^5\mathrm{J} \end{gathered}$$

As only half the potential energy of water is converted into electrical energy,
$\mathrm{Electrical}\mathrm{energy}=\frac{1}{2}\mathrm{P}.\mathrm{E}.=441\times {10}^5\mathrm{J}/\mathrm{hr}$
So, power in watt $\left(\mathrm{J}/\sec \right)=\left(\frac{441\times {10}^5}{60\times 60}\right)$
Therefore, the number of 100 W lamps that can be lit using this energy,
$n=\frac{441\times {10}^5}{3600\times 100}=122.5\approx 122$