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18 January, 21:34

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

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Answers (2)
  1. 18 January, 21:48
    0
    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
  2. 18 January, 21:57
    0
    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
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