Chapter 6.3, Problem 30R

To determine

To calculate: Monthly cost of using a refrigerator.

Answer to Problem 30R

Monthly cost of using the refrigerator is $\underset{\_}{\text{\$42}}$ .

Explanation of Solution

Given:

Time = $\text{10 h}$

Power = $700\text{ W}$

Cost per kWh = $\text{\$0}\text{.20}$

Formula used:

Electrical energy is given by,

  $E=Pt$

where $P$ is power and $t$ is time.

Cost is given by,

  $\text{Cost = (kWh provided)(cost per kWh)}$

Calculation:

To calculate monthly cost, electrical energy used should be calculated first.

Electrical energy is calculated as:

  $\begin{array}{l}E=Pt\\ =(700\text{ W})(10\text{ h})\\ =\left(\frac{700}{1000}\text{ kW}\right)(10\text{ h})\\ =(0.7\text{ kW})(10\text{ h})\\ =7\text{ kWh}\end{array}$

Electrical energy used by the refrigerator is one day = $7\text{ kWh}$

Electrical energy used by the refrigerator is one month = $(7\times 30)\text{ kWh}=210\text{ kWh}$

Cost is calculated as:

  $\begin{array}{l}\text{Cost = (kWh provided)(cost per kWh)}\\ \text{ = (210 kWh)(\$0}\text{.20/kWh)}\\ \text{ = \$42}\end{array}$

Conclusion:

Monthly cost of using the refrigerator is $\underset{\_}{\text{\$42}}$ .