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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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Answer:

See explanation.

Explanation:

Hello,

In this case, we could have two possible solutions:

A) If you are asking for the molar mass, you should use the atomic mass of each element forming the compound, that is copper, sulfur and four times oxygen, so you can compute it as shown below:

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n_{solute}=1mol/L*1.5L=1.5mol

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