Given a mass of 5 kg, an initial temperature of 20 degrees, a final 100 degrees, and a mass of 500 grams of coal, find the efficiency.
mw = 5 kg.
t1 = 20 ° C.
t2 = 100 ° C.
mu = 500 g = 0.5 kg.
C = 4200 J / kg * ° C.
q = 2.7 * 10 ^ 7 J / kg.
Let us write down the definitions for efficiency = Qp * 100% / Qz, where Qp is the useful amount of heat that went to heat the water, Qz is the consumed amount of heat that is released during the combustion of coal.
Qp = C * mw * (t2 – t1).
Qz = q * mu.
The formula for determining the efficiency will take the form: efficiency = C * mw * (t2 – t1) * 100% / q * mu.
Efficiency = 4200 J / kg * ° C * 5 kg * (100 ° C – 20 ° C) * 100% / 0.5 kg * 2.7 * 10 ^ 7 J / kg = 12.4%.
Answer: efficiency = 12.4%.
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