The lightbulb of a refrigerator malfunctions and remains on. The increases in the electricity consumption and cost are to be determined.
Assumptions The life of the lightbulb is more than 1 year.
Analysis The lightbulb consumes 40 W of power when it is on, and thus adds 40 W to the heat load of the refrigerator. Noting that the COP of the refrigerator is 1.3, the power consumed by the refrigerator to remove the heat generated by the lightbulb is
\dot{W}_{\text {refrig }}=\frac{\dot{Q}_{\text {refrig }}}{\mathrm{COP}_{\mathrm{R}}}=\frac{40 \mathrm{~W}}{1.3}=30.8 \mathrm{~W}
Therefore, the total additional power consumed by the refrigerator is
\dot{W}_{\text {total,additional }}=\dot{W}_{\text {light }}+\dot{W}_{\text {refrig }}=40+30.8=70.8 \mathrm{~W}
The total number of hours in a year is
\text { Annual hours }=(365 \text { days } / \mathrm{yr})(24 \mathrm{~h} / \text { day })=8760 \mathrm{~h} / \mathrm{yr}
Assuming the refrigerator is opened 20 times a day for an average of 30 \mathrm{~s}, the light would normally be on for
\begin{aligned}\text { Normal operating hours } &=(20 \text { times } / \text { day })(30 \mathrm{~s} / \text { time })(1 \mathrm{~h} / 3600 \mathrm{~s})(365 \text { days/yr }) \\&=61 \mathrm{~h} / \mathrm{yr}\end{aligned}
Then the additional hours the light remains on as a result of the malfunction becomes
\begin{aligned}\text{Additional operating hours } &= \text{Annual hours} – \text{Normal operating hours}\\&=8760-61=8699 \mathrm{~h} / \mathrm{yr}\end{aligned}
Therefore, the additional electric power consumption and its cost per year are
Additional power consumption ~=\dot{W}_{\text {total, additional }} \times (Additional operating hours)
\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad=(0.0708 \mathrm{~kW})(8699 \mathrm{~h} / \mathrm{yr})=616 \mathrm{kWh} / \mathrm{yr}
and
Additional power cost =( Additional power consumption)(Unit cost)
\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad=(616 \mathrm{kWh} / \mathrm{yr})(\$ 0.12 / \mathrm{kWh})=\$ 73.9 / \mathrm{yr}
Discussion Note that not repairing the switch will cost the homeowner about \$ 75 a year. This is alarming when we consider that at \$ 0.12 / \mathrm{kWh}, a typical refrigerator consumes about \$ 100 worth of electricity a year.
