91影视

First, determine the total electric consumption in a year

$$\begin{gathered} E=Pt \\ E=\left(75.0W\right)\left(1.00yr\right)\left(\frac{3.156脳{10}^{7}s}{1.00s}\right) \\ E=2.367脳{10}^{9}J \end{gathered}$$

The total energy consumption by the bulb in KW-h is

$$\begin{gathered} E=\left(2.367脳{10}^{9}J\right)\left(\frac{1.00KW-h}{3.60脳{10}^{6}J}\right) \\ E=657.5KW-h \end{gathered}$$

So, the total cost in one year is

$$\begin{gathered} m=RE \\ m=\left(0.120\$/KW-h\right)\left(657.5KW-h\right) \\ m=\$78.9 \end{gathered}$$

Therefore, $78.9 is the yearly cost to keep a $75-W$bulb burning day and night.

Total energy consumed by the refrigerator in a year

$$\begin{gathered} E=Pt=\left(400W\right)\left(0.333yr\right)\left(\frac{3.156脳{10}^{7}s}{1.00yr}\right) \\ E=4.208脳{10}^{9}J \\ E=\left(4.208脳{10}^{9}J\right)\left(\frac{1.00KW-h}{3.60脳{10}^{6}J}\right) \\ E=1169KW-h \end{gathered}$$

So, the cost of operating refrigerator for a year

$$\begin{gathered} m=RE \\ m=\left(0.120\$/KW-h\right)\left(1169KW-h\right) \\ m=\$140.3 \end{gathered}$$

Therefore, the yearly cost of operating the refrigerator for a year is $140.3