200 grams of water was heated on an alcohol lamp from 20 degrees to a boil and 10 grams evaporated burned 20 grams of alcohol Determine the efficiency
Data: m1 (mass of heated water) = 200 g = 0.2 kg; t (initial temp.) = 20 ºС, tboil (boiling temperature) = 100 ºС, m2 (mass of evaporated water) = 10 g = 0.01 kg; m3 (mass of alcohol) = 20 g = 0.02 kg.
Constants: C (specific heat capacity) = 4200 J / (kg * K); q (specific heat of combustion of alcohol) = 2.7 * 10 ^ 7 J / kg, L (specific heat of vaporization) = 2300 kJ / kg = 2300000 J / kg.
η = (A / Q) * 100% = (C * m1 * (tboil – t) + L * m2) / (q * m3) = (4200 * 0.2 * (100 – 20) + 2300000 * 0, 01) / (2.7 * 10 ^ 7 * 0.02) = 0.167 = 16.7%.
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