Question

Approximately $\mathbf{5}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathit{k}\mathit{g}$of water falls 50 mover Niagara Falls each second. (a) What is the decrease in the gravitational potential energy of the water–Earth system each second? (b) If all this energy could be converted to electrical energy (it cannot be), at what rate would electrical energy be supplied? (The mass of 1 m³of water is 1000 kg.) (c) If the electrical energy were sold at 1cent/KW.h, what would be the yearly income?

Short answer

a) Decrease in gravitational potential energy of the water earth system each second is$2.7\times {10}^{9}J$

b) Rate of electrical energy supplied is$2.7\times {10}^{9}W$

c) If the electrical energy were sold at 1cent/kWh the early income is$2.4\times {10}^{10}cents$

Step-by-step solution

The given data

a) Mass of the water,$m=5.5\times {10}^{6}kg$

b) Height of Niagara falls, h = 50m

c) Mass of water, $m_{water}=1000kg$

d) Electrical energy sold at, 1cent/kWh

Understanding the concept of gravitational potential energy

We use the concept of gravitational potential energy. We find the change in the gravitational potential energy of the water. From the conservation of energy concept, we can find the rate of electrical energy supplied. Also, we can find energy used in one year and then the cost of energy for one year.

Formulae:

The gravitational potential energy at a height, U = mgh (i)

a) Calculation of the decrease in gravitational potential energy of the water earth system each second

Using the given data in equation (i), the decrease in gravitational potential energy is given as:

$$\begin{gathered} ∆U=5.5\times {10}^6\times 9.8\times 50 \\ =2.7\times {10}^{9}J \end{gathered}$$

Hence, the value of the decrease in energy isrole="math" localid="1661231921857" $2.7\times {10}^{9}J$ .

b) Calculation of the rate of electrical energy supplied

We are given that all energy will convert into electrical energy, so from the energy conservation-principle, gravitational potential energy will convert into electrical energy.

Hence, the rate of electrical energy is$2.7\times {10}^{9}J$ .

c) Calculation of the early income from electrical energy

We know that $1kWh=3.6\times {10}^{6}J$and $1year=3.153\times {10}^{7}s$

We got $2.7\times \frac{{10}^{9}J}{s}$ electrical energy. We can convert it into kWh as follows:

$$\begin{gathered} E=\frac{\left(2.7\times {10}^9\right)J}{s}\times 3.153\times {10}^{7}s\times \left(\frac{1kWh}{3.6\times {10}^{6}J}\right) \\ =2.364\times {10}^{10}kWh \\ \approx 2.4\times {10}^{10}kWh \end{gathered}$$

We havefor.

For$2.4\times {10}^{10}kWh$the cost is$2.4\times {10}^{10}cents$ .

Hence, the required early income is$2.4\times {10}^{10}cents$