Question

A refrigerator using 1005 W runs one-third of the time. How much does the electricity cost to run the refrigerator each month (30 days) at 19¢ per kWh?answer in:____$

Answer 1

We are asked to calculate the cost to run a 1005 W refrigerator over one-third of 30 days.

To do that we will first convert the 1005W into kW using the following conversion factor:

$1kW=1000W$

Multiplying by the conversion factor we get:

$1005W\times\frac{1kW}{1000W}=1.005kW$

Now, we multiply the kW by the number of hours. Since the refrigerator operates one-third of the time during a 30 days period, this means that the time of operation is 10 days. We convert days into hours using the following conversion factor:

$24h=1day$

Multiplying by the conversion factor we get:

$10days\times\frac{24h}{1day}=240h$

Multiplying by the power we get:

$E=(1.005kW)(240h)=241.2kWh$

Now, to determine the cost we multiply by the cost per kWh:

$C=(241.2kWh)(0.19\frac{dollars}{kWh})=45.83$

Therefore, the cost is $45.83