(b) at a school science exhibition, a student designed a model of a hydroelectric power station shown in…

Explanation:

Step1: Identify energy - change in (i)

When the tap is opened, water in the reservoir has gravitational potential energy due to its height. As it falls, this potential energy is converted into kinetic energy. The moving water then rotates the turbine of the generator, converting kinetic energy into electrical energy which lights up the lamp.

Step2: Calculate potential - energy loss per unit time in (ii)1

The formula for gravitational potential energy is $E_p = mgh$. The rate of mass flow of water is $\dot{m}=3\ kg/s$, $g = 9.8\ m/s^2$ and $h = 1.5\ m$. The potential - energy loss per unit time $\dot{E_p}=\dot{m}gh$. $\dot{E_p}=3\times9.8\times1.5=44.1\ J/s = 44.1\ W$

Step3: Calculate efficiency in (ii)2

The power output of the system is the power of the lamp, $P_{out}=20\ W$, and the power input is the potential - energy loss per unit time of the water, $P_{in}=\dot{E_p}=44.1\ W$. The efficiency $\eta=\frac{P_{out}}{P_{in}}\times100%$. $\eta=\frac{20}{44.1}\times100%\approx45.4%$

Answer:

(i) Gravitational potential energy of water in the reservoir is converted into kinetic energy as it falls. Then kinetic energy of water is converted into electrical energy by the generator which lights up the lamp. (ii)1. $44.1\ W$ (ii)2. Approximately $45.4%$