Calculating cost - amps, watts & kwh

[Chalk]

I have this problem and the answer, but I can't get the same answer as the instructor gives.

An energy-efficient water heater draws 12 amps in a standard 115-volt circuit. It costs $75 more than a standard water heater that draws 15 amps. 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?

Answer given by instructor is 2,173.9 hours to break even.

This is what I came up with after searching the internet twice

This is what I calculated:
12amps x 115V = 1380 watts
1380watts x 24 hours/day = 33120 watts
33120 divided by 1000 = 33.12 Kwh
33.12kwh x 0.10cents = $3.312/day
$3.312day divided 24hours = $.138 hour
$75 divided $.138hour = 543.4 hours to break even

I used the process at the bottom of this web page to calculate the answer - http://www.anapsid.org/electricitycost.html

I can't figure out how the answer is 2,173.9 hours

Appreciate any help - nothing more confusing than not knowing what is right and the text book was no help.

 

[skeeter]

(12 A)(115 V)/(1000 w/kW) * (.10 $/kWh) * (t hours) + $75 = (15 A)(115 V)/(1000 w/kW) * (.10 $/kWh) * (t hours)

solve for t.

 

[Deleted member 4993]

I have this problem and the answer, but I can't get the same answer as the instructor gives.

An energy-efficient water heater draws 12 amps in a standard 115-volt circuit. It costs $75 more than a standard water heater that draws 15 amps. 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?

Answer given by instructor is 2,173.9 hours to break even.

This is what I came up with after searching the internet twice

[quote:cbgfwq28]This is what I calculated:
12amps x 115V = 1380 watts
1380watts x 24 hours/day = 33120 watts
33120 divided by 1000 = 33.12 Kwh
33.12kwh x 0.10cents = $3.312/day
$3.312day divided 24hours = $.138 hour
$75 divided $.138hour = 543.4 hours to break even

I used the process at the bottom of this web page to calculate the answer - http://www.anapsid.org/electricitycost.html

I can't figure out how the answer is 2,173.9 hours

Appreciate any help - nothing more confusing than not knowing what is right and the text book was no help.[/quote:cbgfwq28]

electricity savings = 3 * 115 W = 0.345 kW

savings/hr = 0.0345 $/hr

Time break-even = 75/0.0345 hr = ????

 

[Chalk]

I have this problem and the answer, but I can't get the same answer as the instructor gives.

An energy-efficient water heater draws 12 amps in a standard 115-volt circuit. It costs $75 more than a standard water heater that draws 15 amps. 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?

Answer given by instructor is 2,173.9 hours to break even.

This is what I came up with after searching the internet twice

[quote:12gkm7nt]This is what I calculated:
12amps x 115V = 1380 watts
1380watts x 24 hours/day = 33120 watts
33120 divided by 1000 = 33.12 Kwh
33.12kwh x 0.10cents = $3.312/day
$3.312day divided 24hours = $.138 hour
$75 divided $.138hour = 543.4 hours to break even

I used the process at the bottom of this web page to calculate the answer - http://www.anapsid.org/electricitycost.html

I can't figure out how the answer is 2,173.9 hours

Appreciate any help - nothing more confusing than not knowing what is right and the text book was no help.

electricity savings = 3 * 115 W = 0.345 kW

savings/hr = 0.0345 $/hr

Time break-even = 75/0.0345 hr = ????[/quote:12gkm7nt]

Okay I can see that now and I went that way for a second the other day, but stopped for some reason :? . Thank you