Question

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

Answer 1

Answer:

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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 = $\frac{2025}{1} million Joule per second$

P = $\frac{2025}{1}\times 10^{6} Watt$

P = $\frac{2.025}{1}\times 10^{9} Watt$

b.

Input power =  2025 million Watt

Output power = 800 million Watt

Efficiency = output power / input power

Efficiency = 800 / 2025 = 0.395

Efficiency = 39.5%

Answer 2

A) $P = 75 \times (2.7\times10^7) = 2.025\times10^9 W$

b) $Efficiency = \frac{8\times10^8}{2.025\times10^9} \times 100 \approx 39.5\%$