With an electric boiler, 2.5 kg of water can be heated from 20 ° C to boiling in 10 minutes. The efficiency of the boiler is 80%, and during operation it consumed a current of 7 A. Find the voltage in the network.
To find the mains voltage, we will use the formula: η = Apol / Azat = Sv * m * (t2 – t1) / (U * I * t), whence we express: U = Sv * m * (t2 – t1) / (I * t * η).
Constants and variables: Sv – heat capacity of water (Sv = 4200 J / (kg * K)); m is the mass of heated water (m = 2.5 kg); t2 – boiling point (t2 = 100 ºС); t1 – initial temperature (t1 = 20 ºС); I – current strength (I = 7 A); t is the operating time of the electric boiler (t = 10 min = 600 s); η – efficiency (η = 80% = 0.8).
Calculation: I = U = Sv * m * (t2 – t1) / (I * t * η) = 4200 * 2.5 * (100 – 20) / (7 * 600 * 0.8) = 250 V.
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