Problem 18.66 7 of 15 Part A A particular household uses a 2.2-kW heater 1.5 h/d
ID: 1870799 • Letter: P
Question
Problem 18.66 7 of 15 Part A A particular household uses a 2.2-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 2.1 kWh/day If electricity costs $0.105 per kWh, what will be their monthly bill (30 days)? Express your answer using two significant figures. Cost S per month Submit Request Answer Part B How much coal (which produces 7500 kcal/kg) must be burned by a 38 %-efficient power plant to provide the yearly needs of this household? Express your answer using two significant figures. rn = kg Submit Re t AnExplanation / Answer
a)
2.2 kw * 1.5h = 3.3 kwh
100 w = 0.1 kw x 4bulbs = 0.4 kw * 7h = 2.8kwh
2.7 kw * 1 h = 2.7 kwh
And given 2.1 kwh each day.
Now, add all of these kwh values up and that's the total kwh. Now multiply that by 30 days. Now multiplythe total
number of kwh / month by the monthly charge and you haveyour total monthly bill.
0.105 x 30 x (3.3 + 2.8 + 2.7 + 2.1) = $34.335
b)
b) 1 cal = 4.1868 J
7.5 x 10^3 kcal/kg = 7.5 x 10^3 x 4.1868 kJ/kg = 3.14 x 10^4 kJ/kg
every year 365*10.9 kWh = 3978.5 kWh = 3978.5*3600 kJ = 1.4323 x 10^7 kJ
assume x kg,
3.14 x 10^4 x 38% = 1.4323 x 10^7
x = 1.200 x 10^3 kg