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zepelin [54]
3 years ago
7

Compare the amount of carbon dioxide released in one year from burning coal to power 10, 65-watt incandescent bulbs with the amo

unt released from powering 10, 13-watt compact fluorescent light (CFL) bulbs. Assume the bulbs are on four hours per day for 365 days. You will need to determine the kilowatt hours (kWh) used. First, multiply the wattage of the bulbs by the number of light bulbs to determine the total watts used in one hour. Then multiply the result by time in hours to obtain the watt hours. Next, divide the result by 1000 to obtain kilowatt hours. On average, 2.1 pounds of carbon dioxide are released for every kWh of electricity produced.
Chemistry
1 answer:
Ahat [919]3 years ago
8 0

Answer:

Incandescent bulbs: carbon dioxide released ≈ 1993 pounds

CFL bulbs: carbon dioxide released ≈ 399 pounds

Therefore, CFL bulbs release less amount of carbon dioxide as compared to Incandescent bulbs.

Explanation:

Incandescent bulbs:

Total watts = 10*65

Total watts = 650 watts

watt hours in 1 day = 650*4

watt hours in 1 day = 2600 Wh

In one year,

watt hours in 365 days = 2600*365

watt hours in 365 days = 949000 Wh

Convert into kilowatt hours

kilowatt hours in 365 days = 949000/1000

kilowatt hours in 365 days = 949 kWh

2.1 pounds of carbon dioxide is released for 1 kWh

carbon dioxide released = 949*2.1

carbon dioxide released ≈ 1993 pounds

Compact fluorescent light (CFL) bulbs:

Total watts = 10*13

Total watts = 130 watts

watt hours in 1 day = 130*4

watt hours in 1 day = 520 Wh

In one year,

watt hours in 365 days = 520*365

watt hours in 365 days = 189800 Wh

Convert into kilowatt hours

kilowatt hours in 365 days = 189800/1000

kilowatt hours in 365 days = 189.8 kWh

2.1 pounds of carbon dioxide is released for 1 kWh

carbon dioxide released = 189.8*2.1

carbon dioxide released ≈ 399 pounds

Conclusion:

Therefore, CFL bulbs release less amount of carbon dioxide (399 lb) as compared to Incandescent bulbs (1993 lb)

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Answer:

4.4×10² cm³

Explanation:

From the question given above, the following data were obtained:

Diameter (d) = 68.3 mm

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Next, we shall convert the diameter (i.e 68.3 mm) to cm.

This can be obtained as follow:

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68.3 mm = 68.3 mm / 10 mm × 1 cm

68.3 mm = 6.83 cm

Therefore, the diameter 68.3 mm is equivalent 6.83 cm.

Next, we shall convert the height (i.e 0.120 m) to cm. This can be obtained as follow:

1 m = 100 cm

Therefore,

0.120 m = 0.120 m/ 1 m × 100 cm

0.120 m = 12 cm

Therefore, the height 0.120 m is equivalent 12 cm.

Next, we shall determine the radius of the cylinder. This can be obtained as follow:

Radius (r) is simply half of a diameter i.e

Radius (r) = Diameter (d) /2

r = d/2

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Radius (r) =?

r = d/2

r = 6.83/2

r = 3.415 cm

Finally, we shall determine the volume of the cylinder as follow:

Radius (r) = 3.415 cm

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Volume (V) =?

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V = 440 cm³

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Dimensional analysis can only be used for time, distance, mass, speed, volume, and chemical quantities: False.

<h3>What is dimensional analysis?</h3>

A dimensional analysis can be defined as an analysis of the relationships that exist between different physical quantities by identifying their fundamental (base) quantities, especially for the purpose of inferences.

Dimensional analysis is also referred to as unit-factor method or factor-label method and it is typically used to convert a different type of unit to another.

Generally, dimensional analysis can only be used for the following physical quantities:

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In conclusion, a dimensional analysis cannot be used for chemical quantities such as a mole i.e it has no dimensional formula.

Read more on dimensional analysis here: brainly.com/question/24514347

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