Question

Lee watches TV for 4 hours per day. During that time, the TV consumes 250 watts per hour. Electricity costs (18 cents)/(1 kilowatt-hour). How much does Lee's TV cost to operate for a month of 30 days?

Answer 1

The cost of operating Lee's TV for a month of 30 days is $ 5.4.

Explanation:

Power is the rate of change of energy in time. At first we need to determine monthly energy consumption under the assumption that power remains constant in time:

`E_(month) = E_(hour)\cdot t_(day)\cdot t_(month)`

Where:

`E_(month)` - Monthly energy consumption, measured in watt-hours.

`E_(hour)` - Hourly consumption rate, measured in watts per hour.

`t_(day)` - Daily time, measured in hours per day.

`t_(month)` - Number of days within a month, measured in days.

If we know that
`E_(hour) = 250\,Wh`,
`t_(day) = 4\,(h)/(day)` and
`t_(month) = 30\,days`, then:

`E_(month) = \left(250\,(W)/(h) \right)\cdot \left(4\,(h)/(day) \right)\cdot \left(30\,day \right)`

`E_(month) = 30000\,Wh`

A kilowatt-hour equals 1000 watt-hours, then:

`E_(month) = 30\,kWh`

Finally, we get the monthly cost to operate a TV by multiply the result above by electricity unit cost, that is: (1 cent - 0.01 USD)

`C_(month) = c\cdot E_(month)`

Where:

`C_(month)` - Monthly energy cost, measured in US dollars.

`E_(month)` - Monthly energy consumption, measured in kilowatt-hours.

`c` - Electricity unit cost, measured in US dollars per killowatt-hour.

If we know that
`E_(month) = 30\,kWh` and
`c = 0.18\,(USD)/(kWh)`, then:

`C_(month) = (30\,kWh)\cdot \left(0.18\,(USD)/(kWh) \right)`

`C_(month) = 5.4\,USD`

The cost of operating Lee's TV for a month of 30 days is $ 5.4.