Your utility company charges 11 cents per kilowatt-hour of electricity. Complete parts (a) and (b) below. a. What is the daily cost of keeping lit a 75-watt light bulb for 12 hours each day? b. How much will you save in a year if you replace the bulb with an LED bulb that provides the same amount of light using only 20 watts of power?

a. 9.9 cents

b. 26.5 dollars

Step-by-step explanation:

The first thing they mention to us is the cost of energy 11 cents / kwh

a) In the first point they ask us to calculate the daily cost of keeping the flashlight on for 12 hours, they tell us that it consumes 75 watts per hour, but this value must be passed to kW because the cost is given in kWh

to pass it, it would be: 75 watt * 1 kw / 1000 watt = 0.075 kw,

the formula to apply would be like this:

Daily cost = flashlight cost * time * energy cost

we all know them so we replace:

Daily cost = 0.075 kw * 12 h * 11 cents / kwh = 9.9 cents

b) Now, to know the annual savings, we must calculate the difference between the expense of the normal flashlight and the LED, let's calculate the daily cost of the LED, like this:

20 watt * 1 kw / 1000 watt = 0.02 kw

Daily cost = 0.02 kw * 12 h * 11 cents / kwh = 2.64 cents

Now by calculating the difference we multiply by 365 which are the days of a year.

Savings / year = (9.9 - 2.64) * 365 = 2649.9 cents = 26.5 dollars.

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