A 750W toaster and a 1200W electric frying pan are plugged into the same 100v outlet. How much will it cost to operate the two aplliances at 8¢ per kW.h?
My Work:
750W = 0.75kW
1200W = 1.2kW
0.75kW+1.2kW = 1.95kW
8¢/kW.h = $0.08/kw.h
Therefore: 1.95kW.h x $0.08/kW.h = $0.156
Am i right?
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My teacher told us that the answer is $1.56. So please help me.
PS: Where I have converted the 750 and 1200 W into kW (that is correct), but when i calculate the final cost i put 0.75kW.h and 1.2kW.h <<-- is adding the .h at the end like that allow?
2006-11-29 08:19:18 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Physics
You are right, perhaps your teacher calculated for 10 hours of operation.
and technically you are not allowed to multiply by .h unless you want to calculate the cost for a definite time, but it works here because you want to find the cost for one hour (i.e 1.95 kW for One hour gives you the 1.95 kWh)
Remember: kW represents power
kWh represents energy
2006-11-29 08:31:41 · answer #1 · answered by Mysterious 2 · 0⤊ 0⤋
The ? as asked is ambiguous. It doesn't say what time period the appliances operate for. If 1 hr is assumed, then the units come out like this:
$.08/kw.hr * 1.95kw = $0.156/hr
Note how the kw in the first 2 terms cancel, leaving only $/hr
Your teacher is off by a factor of 10; $0.156 = 15.6¢
2006-11-29 16:35:59 · answer #2 · answered by Steve 7 · 0⤊ 0⤋
I think you got it right. It will cost $0.156 per hour to operate both appliances. And I have a degree in physics and have worked as an electrical engineer.
2006-11-29 16:24:49 · answer #3 · answered by campbelp2002 7 · 0⤊ 0⤋
Your answer is correct. It will cost $0.156 per hour to operator the two devices. Your teacher is wrong
2006-11-29 16:23:40 · answer #4 · answered by Anonymous · 0⤊ 0⤋