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The wattage of an appliance indicates the average power consumption in watts (W)(W), where 1 W=1 J/s1 \mathrm{~W}=1 \mathrm{~J} / \mathrm{s}. What is the difference in the number of kJ\mathrm{kJ} of energy consumed per month between a refrigeration unit that consumes 625 W625 \mathrm{~W} and one that consumes 855 W855 \mathrm{~W} ? If electricity costs $0.15\$ 0.15 per kWh\mathrm{kWh}, what is the monthly cost difference to operate the two refrigerators? (Assume 30.030.0 days in one month and 24.024.0 hours per day.)

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Answered 2 years ago
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From the problem we know:

1W=1J/s1\:\texttt{W}=1\:\texttt{J/s} E1=625W=625J/sE_1=625\:\texttt{W}=625\:\texttt{J/s} E2=855W=855J/sE_2=855\:\texttt{W}=855\:\texttt{J/s}
Monthly energy consumption for refrigerator 1:

625J/s30days1month24hours1day60min1hour60s1min==1.62109J/month=1.62106kJ/month\begin{aligned} &625\:\texttt{J/s}\cdot\frac{30\:\texttt{days}}{1\:\texttt{month}}\cdot\frac{24\:\texttt{hours}}{1\:\texttt{day}}\cdot\frac{60\:\texttt{min}}{1\:\texttt{hour}}\cdot\frac{60\:\texttt{s}}{1\:\texttt{min}}=\\ &=1.62\cdot10^9\:\texttt{J/month}=1.62\cdot10^6\:\texttt{kJ/month} \end{aligned}

Monthly energy consumption for refrigerator 2:

855J/s30days1month24hours1day60min1hour60s1min==2.22109J/month=2.22106kJ/month\begin{aligned} &855\:\texttt{J/s}\cdot\frac{30\:\texttt{days}}{1\:\texttt{month}}\cdot\frac{24\:\texttt{hours}}{1\:\texttt{day}}\cdot\frac{60\:\texttt{min}}{1\:\texttt{hour}}\cdot\frac{60\:\texttt{s}}{1\:\texttt{min}}=\\ &=2.22\cdot10^9\:\texttt{J/month}=2.22\cdot10^6\:\texttt{kJ/month} \end{aligned}

The difference in monthly consumption between the refrigerator 1 and 2:

2.22106kJ/month1.62106kJ/month=5.96105kJ/month\begin{aligned} 2.22\cdot10^6\:\texttt{kJ/month}-1.62\cdot10^6\:\texttt{kJ/month}=5.96\cdot10^5\:\texttt{kJ/month} \end{aligned}

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