Given a mass of 5 kg, an initial temperature of 20 degrees, a final 100 degrees

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.

Efficiency -?

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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