The Physics of Everyday Phenomena
The Physics of Everyday Phenomena
8th Edition
ISBN: 9780073513904
Author: W. Thomas Griffith, Juliet Brosing Professor
Publisher: McGraw-Hill Education
Question
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Chapter 1, Problem 3SP

(a)

To determine

The number of hours in one year the light is on.

(a)

Expert Solution
Check Mark

Answer to Problem 3SP

The number of hours in one year the light is on is 1750hr.

Explanation of Solution

Given Info: A bulb is switched on for 5h, for 350 days.

Write the expression for the total time the light is on.

t=number of days×hours the light is on

Here,

t is the total time the light is on

Substitute 5 hr for the hours the light is on and 350 for the number of days to find the total time the light is on.

t=5hr×350=1750hr

Conclusion:

Therefore, the number of hours in one year the light is onis 1750hr.

(b)

To determine

The energy in kilowatt-hour.

(b)

Expert Solution
Check Mark

Answer to Problem 3SP

The energy in kilowatt-hour is 131.250kW-hr.

Explanation of Solution

Given Info: A bulb is switched on for 5h, for 350 days.

Write the expression for the energy.

E=Pt

Here,

E is the energy

P is the power

t is the time

Substitute 75W for P and 1750hr for t to get E.

E=(75W)(1,750hr)=131,250W-hr×1kW-hr1,000W-hr=131.250kW-hr

Conclusion:

Therefore, the energy in kilowatt-hour is 131.250kW-hr.

(c)

To determine

The energy in kilowatt-hour for a 15W bulb.

(c)

Expert Solution
Check Mark

Answer to Problem 3SP

The energy in kilowatt-hour is 26.250kW-hr.

Explanation of Solution

Given Info: A bulb is switched on for 5h, for 350 days.

Write the expression for the energy.

E=Pt

Substitute 15W for P and 1750hr for t to get E.

E=(15W)(1,750hr)=26,250W-hr×1kW-hr1,000W-hr=26.250kW-hr

Conclusion:

Therefore, the energy in kilowatt-hour is 26.250kW-hr.

(d)

To determine

The cost of using 75W bulb.

(d)

Expert Solution
Check Mark

Answer to Problem 3SP

The cost of using 75W bulb is $19.69.

Explanation of Solution

Given Info: The cost of electricity is 15¢/kW-hr.

Write the expression for the total cost.

Cost for 75-W bulb=Energy in kW-hr×cost per kW-hr

Substitute 131.250kW-hr for the energy in kW-hr and 15¢/kW-hr for the cost per kW-hr.

Costfor 75-W bulb=131.250kW-hr×15¢/kW-hr=1968.75¢×1$100¢=$19.6875$19.69

Conclusion:

Therefore, the cost of using 75W bulb is $19.69.

(e)

To determine

The cost of using 15W bulb.

(e)

Expert Solution
Check Mark

Answer to Problem 3SP

The cost of using 15W bulb is $3.94.

Explanation of Solution

Given Info: The cost of electricity is 15¢/kW-hr.

Write the expression for the total cost.

Cost for 15-W bulb=Energy in kW-hr×cost per kW-hr

Substitute 26.250kW-hr for the energy in kW-hr and 15¢/kW-hr for the cost per kW-hr.

Cost for 15-W bulb=26.250kW-hr×15¢/kW-hr=393.75¢×1$100¢=$3.9375$3.94

Conclusion:

Therefore, the cost of using 15W bulb is $3.94.

(f)

To determine

The savings made by using 15W bulb.

(f)

Expert Solution
Check Mark

Answer to Problem 3SP

The savings made by using 15W bulb is $15.75.

Explanation of Solution

Given Info: The cost of electricity is 15¢/kW-hr.

Write the expression for the savings made.

Savings = Cost for75-Wbulb-Cost for15-Wbulb

Substitute $19.69 for the cost for 75-W bulb and $3.94 for the cost for the 15-W bulb.

Savings=$19.6875$3.9375=$15.75

Conclusion:

Therefore, the cost of using 15W bulb is $15.75.

(e)

To determine

The savings made by using twenty 15W bulb.

(e)

Expert Solution
Check Mark

Answer to Problem 3SP

The savings made by using twenty 15W bulb is $315.

Explanation of Solution

Given Info: The cost of electricity is 15¢/kW-hr.

Write the expression for the total savings made.

Total savings=savings from using one 15-W bulb×20

Substitute $15.75 for the savings using one 15-W bulb.

Total savings=$15.75×20=$315

Conclusion:

Therefore, the savings made by using twenty 15W bulb is $315.

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Chapter 1 Solutions

The Physics of Everyday Phenomena

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