Determine the efficiency of a 660 W electric heater if 2 kg of water was heated on it from 20
| September 28, 2021 | education
Determine the efficiency of a 660 W electric heater if 2 kg of water was heated on it from 20 to 100 degrees in 35 minutes.
Efficiency is the ratio of useful power to consumed power.
µ = Apol / Azat * 100%.
Apol = C * m * (t2 – t1) = 4200 * 2 * (100 – 10) = 672000 J = 672 kJ, where C = 4200 J / kg * C is the specific heat capacity of water.
Azat = I ^ 2 * R * t = P * t = 660 * 35 * 60 = 1386000 J = 1386 kJ.
µ = 672/1386 * 100% = 48.49%.
Answer: The efficiency of the electric heater is 48.49%.
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