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*(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*(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(dollars)/(kWh))=45.83`

Therefore, the cost is $45.83