How much does it take to burn oil in a 30% efficiency smelting furnace to bring to the melting point and melt 10 tons of copper? The initial temperature of copper is 25C.
Efficiency = 30%.
mm = 10 t = 10000 kg.
t1 = 25 ° C.
t2 = 1083 ° C.
Cm = 400 J / kg * ° C.
qн = 4.4 * 10 ^ 7 J / kg.
λm = 2.1 * 10 ^ 5 J / kg.
The definition for efficiency will have the form: efficiency = Qpol * 100% / Qpot, where Qpol is the useful amount of heat that goes to heat and melt the shallow, Qpot is the spent heat that is released when burning oil.
Qpol = Cm * mm * (t2 – t1) + λm * mm.
Qpot = qн * mn.
Efficiency = (Cm * mm * (t2 – t1) + λm * mm) * 100% / qn * mn.
mn = (Cm * mm * (t2 – t1) + λm * mm) * 100% / qn * efficiency.
mn = (400 J / kg * ° C * 10000 kg * (1083 ° C – 25 ° C) + 2.1 * 10 ^ 5 J / kg * 10000 kg) * 100% / 4.4 * 10 ^ 7 J / kg * 30% = 480 kg.
Answer: to melt copper, it is necessary to burn oil mn = 480 kg.
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