Question

(a) A particular household uses a 2.2-kW heater 3.1 h/day ("on" time), four 98-W lightbulbs 5.1 h/day, a 3.3-kW electric stove element for a total of 1.6 h/day, and miscellaneous power amounting to 2.1 kWh/day. If electricity costs $0.14 per kWh, what will be their monthly bill (assume 1 month = 30 days)?
  dollars
(b) How much coal (which produces 7000 kcal/kg) must be burned by a 35%-efficient power plant to provide the yearly needs of this household (1 year = 365 days, not 12x30 days)?
kg

Answer 1

We know that, E = P x t

Total energy consumed will be given as :

E_total = [(2.2 kW) (3.1 hr/day)] + [(0.392 kW) (5.1 hr/day)] + [(3.3 kW) (1.6 hr/day)] + [(2.1 kWhr/day)]

E_total = [(6.82 kWhr/day) + (1.9992 kWhr/day) + (5.28 kWhr/day) + (2.1 kWhr/day)]

E_total = 16.1992 kWhr/day

If electricity costs $0.14 per kWhr, then their monthly bill will be given as -

Cost (in dollars) = ($0.14 per kWhr) [(16.1992 kWhr/day) (30 days)]

Cost = $ 68.03

(b) How much coal must be burned by a 35% efficient power plant to provide the yearly needs of this household.

m = [(0.35) (16.1992 kWhr/day) (365 day)] / (7000 kcal/kg)

converting kWhr to J and kcal/kg to J/kg :

m = [(0.35) (58317120 J/day) (365 day)] / (2.9288 x 10⁷ J/kg)

m = [(7450012080 J) / (2.9288 x 10⁷ J/kg)]

m = 254.3 kg