Calculate the efficiency of a fossil-fuel power plant that consumes 380 metric tons of coal each hour to produce useful work at the rate of 720 mw. the heat of combustion of coal (the heat due to burning it) is 28 mj/kg.

24.35 %

Explanation:

m = 380 metric tons = 380 x 10^3 kg , time = 1 hour = 3600 second

Power output = 720 MW = 720 x 10^6 Watt

heat of combustion = 28 MJ/kg = 28 x 10^6 J/kg

Total heat given by coal = mass of coal x heat of combustion of coal

Heat input = 380 x 10^3 x 28 x 10^6 = 1064 x 10^10 J

Power input = Heat input / time = 1064 x 10^10 / 3600 = 2.956 x 10^9 Watt

Efficiency = Output power / Input power = (720 x 10^6) / (2.956 x 10^9)

Efficiency = 0.244 = 24.35 %

when it gains or loses electrons

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