To store the energy produced in 1.5 hour by a 160 MW (160 10^6 W) electric-power plant, how many cubic meters of water will have to be pumped from the lower to the upper reservoir? Assume the upper reservoir is 500 m above the lower and we can neglect the small change in depths within each. Water has a mass of 1000 kg for every 1.0 m^3. Express your answer using two significant figures.
Solve this equation for M, the water mass that must be pumped to a height of H = 500 m.
M g H = 160*10^6 W * 5400 s = 8.64*10^11 Joules
Use g = 9.8 m/s^2
Divide the mass by the density for the required volume.
where would I use the 9.8m/s^2? And the density is which number?
EMily, your question baffles me.
MgH=some number
Mass=some number/gH
Mass/volume=density
Volume=Mass/density
I understand the mass should be divided by the density for the required volume. I don't think that I am using the correct numbers. I came up with a volume of 172800 or 1.7 x 10^5 but the answer is incorrect. Can you please help?!! I did (1000V)(10)(500) = 1.7x 10^5
Use g in the first formula I gave you, to get M. They gave you the density nof water, 1000 kg/m^3.
I get 1.763*10^8 m^3 for the mass and
1.763*10^5 m^3 for the volume
To solve this problem, we first need to calculate the total energy produced by the electric-power plant in 1.5 hours. We can use the formula for power:
Power = Energy / Time
Rearranging the formula, we get:
Energy = Power x Time
Given that the power of the electric-power plant is 160 MW (or 160 x 10^6 W) and the time is 1.5 hours, we can calculate the energy produced:
Energy = 160 x 10^6 W x 1.5 hours
Now, let's convert the energy from watt-hours (W) to joules (J). Recall that 1 watt-hour is equal to 3600 joules:
Energy = 160 x 10^6 W x 1.5 hours x 3600 J/W
Next, we can calculate the gravitational potential energy gained by the water when it is pumped from the lower reservoir to the upper reservoir. The formula for gravitational potential energy is:
Gravitational Potential Energy = Mass x Gravity x Height
Given that the height is 500 m and the mass of water is 1000 kg per 1.0 m^3, we can calculate the gravitational potential energy gained:
Gravitational Potential Energy = Mass x Gravity x Height
= (Mass of water per m^3) x (Volume of water) x Gravity
= (1000 kg/m^3) x (Volume of water) x (9.8 m/s^2)
Now, we equate the energy produced by the power plant to the gravitational potential energy gained by the water:
Energy = Gravitational Potential Energy
160 x 10^6 W x 1.5 hours x 3600 J/W = (1000 kg/m^3) x (Volume of water) x (9.8 m/s^2)
To solve for the volume of water, we can rearrange the formula:
Volume of water = (160 x 10^6 W x 1.5 hours x 3600 J/W) / ((1000 kg/m^3) x (9.8 m/s^2))
Now, let's substitute the given values and calculate the volume of water:
Volume of water = (160 x 10^6 W x 1.5 hours x 3600 J/W) / ((1000 kg/m^3) x (9.8 m/s^2))
After evaluating the expression, we get the volume of water required to store the energy produced by the power plant in 1.5 hours.
Finally, remember to express the answer using two significant figures.