Electrical Power - Cost of kW*hr

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An electric air conditioning unit consuming 15 amps at 134 volts operates for 506.4 hours over 21.1 days in July. The power consumption calculates to 2.01 kW, leading to a total energy usage of 1017.864 kW·hr. The local electricity rate is 9.9 cents per kW·hr, which translates to $0.099. The correct calculation for the monthly cost is $100.768, rounded to $100 when excluding cents. The key takeaway is ensuring the final answer is presented in dollars without cents for accuracy.
calvert11
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Homework Statement


An electric air conditioning unit draws 15 amps of direct current from a 134 V direct voltage source, and is used 24 hours a day during 21.1 days in July. How much will the electricity cost for the month if the local electrical rate is
9.9 /kW· hr?

Answer in units of dollars.
(Note: sigfigs do not matter)

Homework Equations



P = I*\DeltaV

The Attempt at a Solution


This seems like a straight forward question:

P = 15*134 = 2010 W = 2.01 kW
t = 21.1 days * 24 hours = 506.4 hours

P*t = 1017.864 kW*hr

Since units aren't given for 9.9, I tried for both dollars and cents:
1017.864* 9.9 = $10076.85
1017.864*.099 = $100.785

Both answers are wrong. I don't know where I'm making the mistake.
 
Last edited:
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The 9.9 is c/kwh, other than that you are correct, except for the slight typo on the last answer
 
Looks like you might have made an arithmetic error when calculating in cents.

1017.864*0.099 = 100.768
 
Your last answer is correct. They want the answer of dolllars, with no cents. That's why they had such strange numbers (like 21.1 days).
 
berkeman said:
Your last answer is correct. They want the answer of dolllars, with no cents. That's why they had such strange numbers (like 21.1 days).
Thank you so much. Removing cents gives the right answer.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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