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## Re: Practice Test Physics # 4

### Re: Practice Test Physics # 4

4. Given a cost of electricity of \$0.15 per kilowatt-hour (kWh), how much is saved by switching from a 60 watt light bulb to a 25 watt compact fluorescent bulb for 200 hours?

A. Less than \$1
B. Between \$1 and \$5
C. Between \$5 and \$50
D. More than \$50
E.

Explanation:

This is strictly a problem of dimensional analysis (i.e. paying attention to the units to get a mathematical result). Begin by identifying a kilowatt as a thousand watts and a “kilowatt-hour (kWh)” as a kilowatt multiplied by an hour. The power saved would be 60 – 25 = 35 watts

Convert watts to kilowatts: 35 watts x (1 kW/1000 watts) = 0.035 kW

Multiply kW by hours: 0.035 kW x 200 h = 7.0 kWh

Multiply kWh by cost/kWh: 7.0 kWh x \$0.15/kWh = \$0.85 which is less than \$1.

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7*0.15 = 1.05 So isn't the answer B
amishley1421

Posts: 1
Joined: Wed May 29, 2019 1:12 pm

### Re: Practice Test Physics # 4

Did the same thing. I think it's B.
mnlarson4143

Posts: 1
Joined: Wed Oct 23, 2019 5:44 pm