Answer:
a
The efficiency is =97.8%
b
The Energy consumed is [tex]=10512*10^5 J[/tex]
c
The annual cost is = $23.36
Explanation:
from the question we are given
Power supplied = 100 W
Power Produced = 97.8 W
The Efficiency can be evaluated as [tex]\eta = \frac{Power \ Produced}{Power \ Supplied}[/tex]
[tex]\eta = \frac{97.8}{100}*\frac{100}{1}[/tex]
=97.8%
To obtain the energy consumed in a year
We have to known that
1 year = 365 days
Total hours = 8 * 365 = 2920 hours [Since 8 hours is used in one day]
and the 2920 hours obtained = (292× 3600) = 10,512,000 sec
Energy = Power × Time
= (100 × 10512000) W.sec
[tex]=10512*10^5 J[/tex]
Generally
[tex]1Kwh = 36*10^5 J\\[/tex]
So ,Energy in (Kwh) [tex]= \frac{10512*10^5}{36*10^5}[/tex]
[tex]= 292 Kwh[/tex]
Given that the rate should be $0.08/kWh
The Annual cost would be
[tex]=292 Kwh * \frac{0.08}{Kwh}[/tex]
[tex]= 23.36[/tex]
Answer:
(a) efficiency = 2.2495%
(b) Total power consumption
for a whole year = 292kwh
(c) cost = $23.36/kwh
Explanation:
Let He be the heat emitted
Pr be the power rating
and Hl Heat emitted in light
Hl = Pr - He = 100-97.8 = 2.2w
(a) Efficiency = 2.2/97.8 x 100 = 2.2495%
(b) Total hours of consumption for a year is = 8 x 365days = 2920hours
Total power consumption for a year is = 2920 x 100 = 292, 000w = 292kwh
(c) If one 1 kwh costs $0.08
then 292kwh will cost 0.08. x 292 = $23.36/kwh
