How much coal (which produces 6900 kcal/kg) mustbe burned by a 39.7%-efficient p

Question

How much coal (which produces 6900 kcal/kg) mustbe burned by a 39.7%-efficient power plant to provide the yearlyneeds of this household? For the second question I'm a little lost (or a lot, not sureyet) What I did was take each individual power converted it toWatts and multiplied the power by the time of usage afterconverting the time to seconds. Then I added all the Energy ( nowin Joules) and I got 5.76e7J. Next I calculated the amount of energy needed to heat thepower plant by taking 39.7% of x = 5.76e7J. I got 1.45e8 J,converted it to Kcal and divided by the conversion given to get kg.I got 5.09 kg but that is incorrect. Where am I going wrong? You are correct.Computer's answer now shown above.
Your receipt is 164-7849 PreviousTries

Explanation / Answer

How much coal (which produces 6900 kcal/kg) mustbe burned by a 39.7%-efficient power
plant to provide the yearly needs of this household? answer:
energy consumed per year = 12 x 480.39 kWh = 5764.68 kWh =2.075 x 1010 J = 4.96 x 106 Kcal
since 1 kWh = 1000 (J/s) x (3600) s = 3.6 x 106 J
if you need x kg of coal to produce this,
x (6900) (.397) = 4.96 x 106
x = 1810.68 kg

hope this helps!

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.