A particular household uses a 2.4-kW heater 1.5 h/day (\"on\" time), four 100-W
ID: 1878783 • Letter: A
Question
A particular household uses a 2.4-kW heater 1.5 h/day ("on" time), four 100-W lightbulbs 7.0 h/day , a 2.7-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)?
Express your answer using two significant figures.
B.)How much coal (which produces 7500 kcal/kg) must be burned by a 31 % -efficient power plant to provide the yearly needs of this household?
Express your answer using two significant figures.
Explanation / Answer
Amount energy used per day = 2.4*1.5 + 4*0.1*7 + 2.7*1 + 2.3
= 11.4 kWh
amount energy used in a month = 11.4*30
= 342 kWh
A) so, monthly bill = 342*0.105
= 36 $ <<<<<<<<<<<------------------Answer
B) let Q is the amount of energy burned.
0.31*Q = 342 kWh
0.31*Q = 342*3.6*10^6 J
Q = 342*3.6*10^6/0.31
= 3.97*10^9 J
= 3.97*10^9/(4.18) cal
= 9.50*10^8 cal
so, amount coal burned, m = 9.50*10^8/(7500*10^3) kg
= 130 kg <<<<<<<<<<<------------------Answer
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