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:jjExplanation:Energy obtained by burning of 1 kg coal = 27 million Joulea.Energy obtained by burning of 75 kg coal = 27 x 75 = 2025 million Joule Power = Energy obtained per second Power = [tex]\frac{2025}{1} million Joule per second[/tex]P = [tex]\frac{2025}{1}\times 10^{6} Watt[/tex]P = [tex]\frac{2.025}{1}\times 10^{9} Watt[/tex]b.Input power =  2025 million WattOutput power = 800 million WattEfficiency = output power / input power Efficiency = 800 / 2025 = 0.395Efficiency = 39.5%

a) [tex]P = 75 \times (2.7\times10^7) = 2.025\times10^9 W[/tex]b) [tex]Efficiency = \frac{8\times10^8}{2.025\times10^9} \times 100 \approx 39.5\%[/tex]