Question

At steady-state, a porch light with a 100 W incandescent bulb produces 97.8 W of heat energy, the rest of its output being visible light.a) What is the efficiency of the light bulb? (use the definition from the class notes)b) How much energy does the bulb consume annually if it is on for an average of 8 hours per day (in kWh)?c) What is the annual cost of the energy at a rate of $0.08/kWh?

Answer 1

(a) efficiency = 2.2495%

(b) Total power consumption

for a whole year = 292kwh

(c) cost = $23.36/kwh

Step-by-step 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

Answer 2

a

The efficiency is =97.8%

b

The Energy consumed is
`=10512*10^5 J`

c

The annual cost is = $23.36

Step-by-step explanation:

from the question we are given

Power supplied = 100 W

Power Produced = 97.8 W

The Efficiency can be evaluated as
`\eta = (Power \ Produced)/(Power \ Supplied)`

`\eta = (97.8)/(100)*(100)/(1)`

=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

`=10512*10^5 J`

Generally

`1Kwh = 36*10^5 J\\`

So ,Energy in (Kwh)
`= (10512*10^5)/(36*10^5)`

`= 292 Kwh`

Given that the rate should be $0.08/kWh

The Annual cost would be

`=292 Kwh * (0.08)/(Kwh)`

`= 23.36`