A particular household uses a 2.2-kW heater 1.5 h/day ("on" time), four 75-W lightbulbs 5.0 h/day , a 2.5-kW electric stove element for a total of 1.0 h/day , and miscellaneous power amounting to 2.3 kWh/day.

a. If electricity costs $0.105 per kWh, what will be their monthly bill (30 days)?

b. How much coal (which produces 7500 kcal/kg) must be burned by a 37 % -efficient power plant to provide the yearly needs of this household?

Explanation:

Energy consumed per day = (2.2 x 1.5 + 4 x .075 x 5 + 2.5 x 1 + 2.3 )kWh

= 3.3 + 1.5 + 2.5 + 2.3

= 9.6 kWh

Energy consumed in a month = 9.6 x 30

= 288 kWh

cost = 288 x .105

= $ 30.24

b ) 288 kWh = 288 x 60 kW

= (288 x 60)/ 4.2 kcal

= 4114.28 kcal

coal required per month = ( 4114.28 / 7500) x (1 / .37 )

= 1.48 kg

coal required per month= 1.48 x 12

=17.76  kg

=

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