1) In 2011, the total of all residential customers in Halton Region used 54 540 000 m3 of water. Halton Region has a population of around 500 000 people and Milton (part of Halton Region) has a population of around 85 000 people. Assuming all people in Halton use the same amount of water, how much water was used in Milton in 2011?

2) 1 cm3 = 1 mL. The density of water is 1 g/cm3. Approximately what mass of water Milton use in 2011?

3) Most houses in Milton draw their water from the system that draws its water from Lake Ontario. Lak Ontario is 74 m above sea level and Milton is 221 m above sea level. The majority of the water in Milton passes through the water tower on Steeles Ave. It is 55 m from ground to top. In total, how much vertical distance must the water rise from Lake Ontario to the water tower?

4) How much gravitational potential energy does each kg of water gain travelling from Lake Ontario to the top of the water tower?

5) How much energy was needed to bring all the water used in Milton in 2011 from Lake Ontario to the water tower?

6) The energy used to lift the water comes from an electric water pump. Assuming the pump is 85% efficient at lifting the water, how much input electrical energy is needed for a year to lift the water?

7) Much of the energy used by the pump comes from natural gas fired plants (like the new one on Steeles Ave.). A typical efficiency for these power plants is 50%.
a. How much input energy (in the form of chemical energy) from natural gas is needed for a year to lift the water?

b. The chemical energy released from natural gas is around 35 MJ (Mega-Joules) per m3. What volume of gas needs to be burned to pump a year’s worth of water to Milton?

helppp,thats a very long question and its all related so should be easy enough for you guys to help out

See previous post.

Sure, I can help you with your questions. Let's take them one by one.

1) To find out how much water was used in Milton in 2011, we need to calculate the proportion of Milton's population to the total population of Halton.

Population of Milton / Total Population of Halton = Proportion of Milton's population

85,000 / 500,000 = 0.17

Now, we can calculate the amount of water used in Milton by multiplying the total water usage in Halton by the proportion of Milton's population:

Water used in Milton = Total water usage in Halton * Proportion of Milton's population

Water used in Milton = 54,540,000 m3 * 0.17

Therefore, approximately 9,268,800 m3 of water was used in Milton in 2011.

2) To calculate the mass of water, we can use the given density of water, which is 1 g/cm3.

Mass of water = Volume of water * Density of water

Since 1 cm3 = 1 mL, we can directly use the volume of water used in Milton in 2011 (9,268,800 m3) as the volume of water:

Mass of water = 9,268,800 m3 * 1000 g/m3

Therefore, approximately 9,268,800,000 grams (or 9,268,800 kilograms) of water was used in Milton in 2011.

3) To calculate the vertical distance the water must rise from Lake Ontario to the water tower, we need to subtract the elevation of Lake Ontario from the elevation of Milton, taking into account the height of the water tower.

Vertical distance = (Elevation of Milton + Height of water tower) - Elevation of Lake Ontario

Vertical distance = (221 m + 55 m) - 74 m

Therefore, the vertical distance the water must rise is 202 meters.

4) The gravitational potential energy gained by each kilogram of water traveling from Lake Ontario to the top of the water tower can be calculated using the formula:

Potential Energy = Mass of water * Gravitational acceleration * Vertical distance

Gravitational acceleration is approximately 9.8 m/s2.

Potential Energy = 1 kg * 9.8 m/s2 * 202 m

Therefore, each kilogram of water gains approximately 1,982.8 Joules of gravitational potential energy.

5) To calculate the total energy needed to lift all the water used in Milton in 2011, we multiply the potential energy gained per kilogram of water by the mass of water used in Milton.

Total Energy = Potential Energy per kg * Mass of water used in Milton

Total Energy = 1,982.8 J/kg * 9,268,800 kg

Therefore, the total energy needed to lift all the water used in Milton in 2011 is approximately 18,364,294,400 Joules.

6) To calculate the input electrical energy needed to lift the water, we need to take into account the efficiency of the water pump.

Input Electrical Energy = Total Energy / Efficiency

Input Electrical Energy = 18,364,294,400 J / 0.85

Therefore, approximately 21,606,702,823 Joules of input electrical energy is needed for a year to lift the water.

7a) To calculate the input energy in the form of chemical energy from natural gas needed to lift the water, we need to take into account the efficiency of the power plants.

Input Energy from Natural Gas = Input Electrical Energy / Efficiency of power plants

Input Energy from Natural Gas = 21,606,702,823 J / 0.50

Therefore, approximately 43,213,405,646 Joules of input energy from natural gas is needed for a year to lift the water.

7b) To determine the volume of gas burned to pump a year's worth of water to Milton, we need to divide the total energy needed by the energy released per cubic meter of natural gas burned.

Volume of gas = Input Energy from Natural Gas / Chemical energy released per m3

Volume of gas = 43,213,405,646 J / 35,000,000 J/m3

Therefore, approximately 1,234.10 m3 of gas needs to be burned to pump a year's worth of water to Milton.

I hope this helps! Let me know if you have any further questions.