Question

Some electric power companies use water to store energy.Water is pumped from a low reservoir to a high reservoir.To store the energy produced in 1.0 hour by a 180-MWelectric power plant, how many cubic meters of water willhave to be pumped from the lower to the upper reservoir?Assume the upper reservoir is an average of 380 m abovethe lower one. Water has a mass of for every 1.00 x 10^3 kg for every 1.0m^3.

Answer 1

Answer:

1.74 x 10⁵ m³

Explanation:

The power is equal to:

$P=\frac{E}{t}\rightarrow E=Pt$

Where P is the power, E is the energy and t is the time. So, we can calculate the energy replacing

P = 180 MW = 180 x 10⁶ W

t = 1 hour = 3600 s

So, the energy is

$\begin{gathered} E=(180\times10^6\text{ W\rparen\lparen3600 s\rparen} \\ E=6.48\times10^{11}\text{ J} \end{gathered}$

On the other hand, the energy stored will be potential, so

$E=mgh$

Where m is the mass, g is the gravity and h is the height. Solving for m and replacing E = 6.48 x 10^11 J, g = 9.8 m/s², and h = 380 m, we get:

$\begin{gathered} m=\frac{E}{gh} \\ \\ m=\frac{6.48\times10^{11}\text{ J}}{9.8\text{ m/s}^2\text{ \lparen380 m\rparen}} \\ \\ m=1.74\times10^8\text{ kg} \end{gathered}$

Finally, we can calculate the volume of the water in cubic meters using the given ratio that Water has a mass of 1.00 x 10^3 kg for every 1.0m^3, so

$1.74\times10^8\text{ kg }\times\frac{1\text{ m}^3}{1\times10^3\text{ kg}}=1.74\times10^5\text{ m}^3$

Therefore, the answer is 1.74 x 10⁵ m³