An energy-efficient water heater draws 4 amps in a 220-volt circuit. It costs $75 more that a standard water heater that draws 1
8 amps in a standard 115-volt circuit. If electricity costs 10 cents per kilowatt-hour, how long would you have to run the efficient water heater to recoup the difference in price? Show your calculations.
The answer to the question above is "630.25 hours" which is the time to recoup the difference in the water heater cost. This problem can be solved using a simple algebra equation which stated as (18*115*0.1/1000)x - (4*220*0.1/1000)x = $75. In this formula, x represents the time to recoup the difference in the water heater cost (Calculation: 0.207x-0.088x=75 --> 0.119x = 75 --> x = 630.25)<span>. </span>
We simplify the cross product first, thereafter the solution of the cross product is now simplified with the dot product as shown in the step by step calculation in the attachment
