## Related questions with answers

The wattage of an appliance indicates the average power consumption in watts $(W)$, where $1 \mathrm{~W}=1 \mathrm{~J} / \mathrm{s}$. What is the difference in the number of $\mathrm{kJ}$ of energy consumed per month between a refrigeration unit that consumes $625 \mathrm{~W}$ and one that consumes $855 \mathrm{~W}$ ? If electricity costs $\$ 0.15$ per $\mathrm{kWh}$, what is the monthly cost difference to operate the two refrigerators? (Assume $30.0$ days in one month and $24.0$ hours per day.)

Solution

VerifiedFrom the problem we know:

$1\:\texttt{W}=1\:\texttt{J/s}$
$E_1=625\:\texttt{W}=625\:\texttt{J/s}$
$E_2=855\:\texttt{W}=855\:\texttt{J/s}$

Monthly energy consumption for refrigerator 1:

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

Monthly energy consumption for refrigerator 2:

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

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

$\begin{aligned} 2.22\cdot10^6\:\texttt{kJ/month}-1.62\cdot10^6\:\texttt{kJ/month}=5.96\cdot10^5\:\texttt{kJ/month} \end{aligned}$

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