Problem 109P

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.)

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