Question

A power station burns 75 kilograms of coal per second. Each kg of coal contains 27 million joules of energy.

a. What is the total power of this power station in watts? (watts=joules/second)
b. The power station’s output is 800 million watts. How efficient is this power station?

Answer 1

a)
`P = 75 * (2.7*10^7) = 2.025*10^9 W`

b)
`Efficiency = (8*10^8)/(2.025*10^9) * 100 \approx 39.5\%`

Answer 2

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Step-by-step explanation:

Energy obtained by burning of 1 kg coal = 27 million Joule

a.

Energy obtained by burning of 75 kg coal = 27 x 75 = 2025 million Joule

Power = Energy obtained per second

Power =
`(2025)/(1) million Joule per second`

P =
`(2025)/(1)* 10^(6) Watt`

P =
`(2.025)/(1)* 10^(9) Watt`

b.

Input power = 2025 million Watt

Output power = 800 million Watt

Efficiency = output power / input power

Efficiency = 800 / 2025 = 0.395

Efficiency = 39.5%