Question

Consider a refrigerator that consumes 320 W of electric power when it is running. If the refrigerator runs only one quarter of the time and the unit cost of electricity is $0.09/kWh, the electricity cost of this refrigerator per month (30 days) is
A. $3.56
B. $5.18
C. $8.54
D. $9.28
E. $20.74

Answer 1

Answer:

B. $5.18

Explanation:

The power of refrigerator is given:

Power = P = 320 W = 0.320 KW

Since, the electricity cost of the month is required, therefore the no. of hours in a month have to be calculated. Therefore;

No. of hours in a month = (1 month)(30 days/month)(24 h / day)

No. of hours in a month = 720 hours

It is given that the refrigerator runs only one-quarter of the time, thus the time of electric consumption will be:

Time of electric consumption = (1/4)(No. of hours in a month)

Time of electric consumption = (1/4)(720 hours)

Time of electric consumption = 180 hours

Now, we will compute the electrical energy used, as follows:

Electric Energy Consumed = (Power)(Time of electric consumption)

Electric Energy Consumed = (0.32 KW)(180 h)

Electric Energy Consumed = 57.6 KWh

Finally, we calculate the electricity cost per month as follows:

Monthly Cost = (Unit Cost)(Electric Energy Consumed)

Monthly Cost = ($ 0.09/KWh)(57.6 KWh)

Monthly Cost = $5.18

Answer 2

Answer:

B. $5.18

Explanation:

Cost of electricity per kWh = $0.09

Power consumption of refrigerator = 320W = 320/1000 = 0.32kW

In a month (30 days) the refrigerator works 1/4 × 30 days = 7.5 days = 7.5 × 24 hours = 180 hours

Energy consumed in 180 hours = 0.32kW × 180h = 57.6kWh

Cost of electricity of 57.6kWh energy consumed by the refrigerator = 57.6 × $0.09 = $5.18