### Re: Practice Test Physics # 4

Posted:

**Wed May 29, 2019 2:46 pm**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

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