A standard 60 W light bulb has a voltage of 130 volts. So, we use this conversion, the Faraday's constant which is equal to approximately 96,500 Coulombs per mole electron, and the Avogadro's number equal to 6.022×10²³ particles/mole . The solution is as follows:
W = Energy/time
60 W = x J/1 s
x = 60 J = 60 C·V
(60 C·V)*(1/130 V)*(1 mole e/96,500 C)*(6.022×10²³ electrons/mole electron)
= 2.88×10¹⁸ electrons
Answer:
QC = 122 KJ
QH = 2.64 x 122 = 322 KJ
Explanation:
TH = 500 Degree C = 500 + 273 = 773 K
TC = 20 degree C = 20 + 273 = 293 K
W cycle = 200 KJ
Use the formula for the work done in a cycle
Wcycle = QH - QC
200 = QH - QC ..... (1)
Usse
TH / TC = QH / QC
773 / 293 = QH / QC
QH / QC = 2.64
QH = 2.64 QC Put it in equation (1)
200 = 2.64 QC - QC
QC = 122 KJ
So, QH = 2.64 x 122 = 322 KJ
The horizontal component is 2.0 cos(30°) = 1.732 m/s²
The vertical component is 2.0 sin(30°) = 1 m/s²
It will take 2.77..... seconds
