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The wattage of an appliance indicates its average power consumption in watts (W), where

Chapter 3, Problem 109P

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QUESTION:

The wattage of an appliance indicates its average power consumption in watts (W), where 1 W = 1 J/s. What is the difference in the number of kJ of energy consumed per month between a refrigeration unit that consumes 625 W and one that consumes 855 W? If electricity costs $0.15 per kWh, what is the monthly cost difference to operate the two refrigerators? (Assume 30.0 days in one month and 24.0 hours per day.)

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QUESTION:

The wattage of an appliance indicates its average power consumption in watts (W), where 1 W = 1 J/s. What is the difference in the number of kJ of energy consumed per month between a refrigeration unit that consumes 625 W and one that consumes 855 W? If electricity costs $0.15 per kWh, what is the monthly cost difference to operate the two refrigerators? (Assume 30.0 days in one month and 24.0 hours per day.)

ANSWER:

Solution 109P

Here, we are going to calculate the monthly cost difference to operate the two refrigerator.

Step 1:

Calculation of the energy consumed by each refrigerator per month.

Given

For refrigerator , energy consumed = 625 W

For refrigerator , energy consumed = 855 W

Electricity cost = $0.15/ kWh

We know,

kilowatt = kilowatt-hour / hour

P = E / t

watt = joule / second

W = J / s

And 1 kWh = 1000 W x (60 x 60 )s

        = 36 x 105 Ws = 36 x 105 J

1 W = 1 J/s

625 W = 625 W x  = 625 J/s

855 W = 855 W x  = 855 J/s

For refrigerator 1 , energy consumed = 625 W = 625 J/s

For refrigerator 2 , energy consumed = 855 W = 855 J/s

Thus energy consumed per month by refrigerator 1

 = 625 J/s x  x  x x

= 1620000000 J/month

= 162 x 104 kJ

Similarly,

energy consumed per month by refrigerator 2

                        855 W x  x  x  x x

                        = 2216160000 J/month

                        = 221.6 x 104 kJ

Thus, difference in energy consumed by both the refrigerator

= (221.6 x 104 kJ -162 x 104 kJ)

= 59.6 x 104 kJ

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