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Calculating yearly cost of running coffee makers
Find the total cost for operating two 15Ω coffee makers connected to a 120V power supply for two hours a day each for one year.
##R = 15Ω##
##V = 120V##
##∆t=(2h)(365)= 730h## Two hours each 365 days a year.
##r_¢=6.4¢/kWh##
##I = V/R = 8A##
There is some basic physics involved, it's the answer I'm concerned with arithmetically.
I find the power of one coffee maker using :
##P = IV = (8A)(120V) = 0.96kW##
There are two coffee makers, so the total power is :
##P_{Total} = (2)(0.96kW) = 1.92kW##
The energy would then be :
##E = P_{Total}∆t = (1.92 kW)(730h) = 1401.6 kWh##
Therefore the total cost is :
##C = Er_¢ = (1401.6 kWh)(6.4¢/kWh) = 8970.24¢ = $89.70##
Is this correct? The answer is listed as $370.02 so I'm not so sure.
Homework Statement
Find the total cost for operating two 15Ω coffee makers connected to a 120V power supply for two hours a day each for one year.
Homework Equations
##R = 15Ω##
##V = 120V##
##∆t=(2h)(365)= 730h## Two hours each 365 days a year.
##r_¢=6.4¢/kWh##
##I = V/R = 8A##
There is some basic physics involved, it's the answer I'm concerned with arithmetically.
The Attempt at a Solution
I find the power of one coffee maker using :
##P = IV = (8A)(120V) = 0.96kW##
There are two coffee makers, so the total power is :
##P_{Total} = (2)(0.96kW) = 1.92kW##
The energy would then be :
##E = P_{Total}∆t = (1.92 kW)(730h) = 1401.6 kWh##
Therefore the total cost is :
##C = Er_¢ = (1401.6 kWh)(6.4¢/kWh) = 8970.24¢ = $89.70##
Is this correct? The answer is listed as $370.02 so I'm not so sure.
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