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sladkih [1.3K]
2 years ago
10

On a website devoted to answering engineering questions, viewers were invited to determine how much power a 100-MW power plant g

enerates annually. The answer declared to be best was submitted by a civil engineering student, who stated, "It produces 100 MW/hr so over the year that's 100*24*365.25 & do the math." a. Carry out the calculation, showing all the units. Answer b. What is wrong with the statement of the question? c. Why was the student wrong in saying that the plant produces 100 MW/hr?
Chemistry
1 answer:
xeze [42]2 years ago
3 0

Answer:

(a) The plant generates 3,153,600,000 MJ a year.

(b) The problem with the question is that "power" is not "generated". What is generated is Energy, and Power is the rate of generation of Energy.

(c) The student is wrong because the plants generates 100 MJ of energy a second. "MW/hr" is not a unit of energy or power, so it has no sense.

Maybe he get confused with MW-h, which is a unit of energy, used to measure electrical consumption.

Explanation:

(a) The power of the plant (100 MW) is the rate at which electrical energy is produced. It has units of [energy]/[time].

In this case, the plant produces 100 MJ/s. The energy produced can also be expressed in other units, like MJh.

To calculate the energy generated in one year, we have

Energy = Power * Time  = \\\\Energy = 100 MW*1year\\\\Energy = (100 \frac{MJ}{s})*(1 year*\frac{365 days}{1year}*   \frac{24hours}{1day}*\frac{3600s}{1hour})\\\\Energy=(100 \frac{MJ}{s})*(31,536,000s)=3,153,600,000MJ

The plant generates 3,153,600,000 MJ a year.

(b) The problem with the question is that "power" is not "generated". What is generated is Energy, and Power is the rate of generation of Energy.

(c) The student is wrong because the plants generates 100 MJ of energy a second. "MW/hr" is not a unit of energy or power, so it has no sense.

Maybe he get confused with MW-h, which is a unit of energy, used to measure electrical consumption. That unit represents the amount of energy consumed or generated by a 1 MW unit in one hour. It is equivalent to 3600 MJ (1 MW-h = 3600 MJ).

In this unit, the 100 MW plant generates 876,000 MWh.

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In the top row labeled Before Collision, the left billiard ball labeled 3 meters per second approaches the right billiard ball l
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Answer:

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Explanation:

Sample Response since i just took it on Edg.

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The symbol P represents the element phosphorus. This element’s atomic number is 15. How many protons and electrons are in a P–3
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Just Lemons Lemonade Recipe Equation:
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Answer:

Explanation:

Hello!

<em>Complete text:</em>

<em>Honors Stoichiometry Activity WorksheetInstructions: </em>

<em>Activity Two: Just Lemons, Inc. Production</em>

<em>Here's a one-batch sample of Just Lemons lemonade production. Determine the percent yield and amount of leftover ingredients for lemonade production and place your answers in the data chart.</em>

<em>Hint: Complete stoichiometry calculations for each ingredient to determine the theoretical yield. Complete a limiting reactant-to-excess reactant calculation for both excess ingredients. </em>

<em>Water 946.36 g </em>

<em>Sugar 196.86 g </em>

<em>Lemon Juice 193.37 g </em>

<em>Lemonade 2050.25g</em>

<em>Leftover Ingredients?</em>

<em>Just Lemons Lemonade Recipe Equation:</em>

<em>2 water + sugar + lemon juice = 4 lemonade</em>

<em>Mole conversion factors:</em>

<em>1 mole of water = 1 cup = 236.59 g</em>

<em>1 mole of sugar = 1 cup = 225 g</em>

<em>1 mole of lemon juice = 1 cup = 257.83 g</em>

<em>1 mole of lemonade = 1 cup = 719.42 g</em>

You have the information on the ingredients used to produce one batch of lemonade and the amount of lemonade produced. To determine which ingredients be leftovers, you have to determine first, which one is the limiting reactant, i.e. the ingredient that will be used up first.

According to the recipe, to make 4 moles of lemonade, you use 2 moles of water, one mole of sugar and one mole of lemon juice, expressed in grams:

2 water  + sugar + lemon juice = 4 lemonade

2*(236.59) + 225g + 257.83g  = 4*(719.42)g

    473.18g + 225g + 257.83g = 2877.68g

So for every 2877.68g of lemonade made, they use 473.18g of water, 225g of sugar, and 257.83g of lemon juice.

You know that they made a batch of 2050.25g, so to detect the limiting reactant, first, you have to calculate, in theory, how much of each ingredient you need to make the given amount of lemonade:

Use cross multiplication

<u>Water:</u>

2877.68g lemonade → 473.18g water

2050.25g lemonade → X= (2050.25*473.18)/2877.68= 337.12g water

Following the recipe, to elaborate 2050.25g of lemonade, you need to use 337.12g of water.

<u>Sugar:</u>

2877.68g lemonade → 225g sugar

2050.25g lemonade → X= (2050.25*225)/2877.68= 160.30g sugar.

To elaborate 2050.25f of lemonade you need to use 160.30g of sugar.

<u>Lemon juice:</u>

2877.68g lemonade → 257.83g lemon juice

2050.25g lemonade → X= (2050.25*257.83)/2877.68= 183.69g lemon juice.

To elaborate 2050.25f of lemonade you need to use 183.69g lemon juice.

Available ingredients vs. theoretical yields for 2050.25g of lemonade:

Water 946.36 g → 337.12g

Sugar 196.86 g → 160.30g

Lemon Juice 193.37 g → 183.69g

The lemon juice will be the first ingredient to be used up, there will be a surplus of water and sugar.

I hope this helps!

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Using above equation as:

\frac{2.12}{0.120}=\frac{3.12}{n_2}

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n_2=\frac{0.3744}{2.12}

n_2=0.17660

n₂ = 0.17660 moles

Molar mass of methane gas = 16.05 g/mol

So, Mass = Moles*Molar mass = 0.17660 * 16.05 g = 2.83 g

<u>2.83 g  are in the piston.</u>

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