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gregori [183]
3 years ago
8

In a certain city, electricity costs $0.15 per kW·h. What is the annual cost for electricity to power a lamp-post for 6.00 hours

per day with (a) a 100.-watt incandescent light bulb (b) an energy efficient 25-watt fluorescent bulb that produces the same amount of light? Assume 1 year = 365 days.
Chemistry
1 answer:
Vanyuwa [196]3 years ago
7 0

(a) Power of bulb is 100 W, converting this into kW.

1 W=\frac{1}{1000}kW

Thus,

100 W =\frac{100}{1000}kW=0.1 kW

The bulb is used for 6 hours per day for a year, in 1 year there are 365 days thus, total hours will be:

t=6\times 365=2190 h

Electricity used will depend on power and number of hours as follows:

E=P\times t=0.1 kW\times 2190 h=219 kW.h

The cost of electricity is $0.15 per kW.h thus, cost of electricity for 219 kW.h will be:

Cost=\$ (219\times 0.15)=\$ 32.85

Therefore, annual cost of incandescent light bulb is \$ 32.85

(b) Power of bulb is 25 W, converting this into kW.

1 W=\frac{1}{1000}kW

Thus,

100 W =\frac{25}{1000}kW=0.025 kW

The bulb is used for 6 hours per day for a year, in 1 year there are 365 days thus, total hours will be:

t=6\times 365=2190 h

Electricity used will depend on power and number of hours as follows:

E=P\times t=0.025 kW\times 2190 h=54.75 kW.h

The cost of electricity is $0.15 per kW.h thus, cost of electricity for 54.75 kW.h will be:

Cost=\$ (54.75\times 0.15)=\$ 8.21

Therefore, annual cost of fluorescent bulb is \$ 8.21.

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Why is an elements atomic mass not listed as a whole number in the periodic table
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A 14.4-gg sample of granite initially at 86.0 ∘C∘C is immersed into 24.0 gg of water initially at 25.0 ∘C∘C. What is the final t
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Answer:

The final temperature of both substances when they reach thermal equilibrium is 31.2 °C

Explanation:

Step 1: Data given

Mass of sample granite = 14.4 grams

Initial temperature = 86.0 °C

Mass of water = 24.0 grams

The initial temperature of water = 25.0 °C

The specific heat of water = 4.18 J/g°C

The specific heat of granite = 0.790 J/g°C

Step 2: Calculate the final temperature

Heat lost = heat gained

Qgranite = - Qwater

Q = m*c*ΔT

m(granite)*c(granite)*ΔT(granite) = -m(water)*c(water)*ΔT(water)

⇒with m(granite) = the mass of granite = 14.4 grams

⇒with c(granite) = The specific heat of granite = 0.790 J/g°C

⇒with ΔT⇒(granite) = the change of temperature of granite = T2 - T1 = T2 - 86.0 °C

⇒with m(water) = the mass of water = 24.0 grams

⇒with c(water) = The specific heat of water = 4.18 J/g°C

⇒with ΔT(water) = the change of temperature of granite = T2 - T1 = T2 -25.0°C

14.4 grams * 0.790 * (T2 - 86.0°C) = -24.0 *4.18 * (T2 - 25.0°C)

11.376T2 - 978.336 = -100.32T2 + 2508

111.696 T2 = 3486.336

T2 = 31.2 °C

The final temperature of both substances when they reach thermal equilibrium is 31.2 °C

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