Answer :
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]
b) [tex]Efficiency = \frac{8\times10^8}{2.025\times10^9} \times 100 \approx 39.5\%[/tex]
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 = [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 Watt
Output power = 800 million Watt
Efficiency = output power / input power
Efficiency = 800 / 2025 = 0.395
Efficiency = 39.5%