An electric kettle for an operating voltage of 220 V has a heater with a resistance of 22.0 Ohm. The efficiency of the kettle is 95%. How much water can be brought to a boil from an initial temperature of 10 degrees in 10 minutes. Disregard the heat capacity of the kettle.
To find out the mass of water brought to a boil, we use the formula: ηech = Sv * mw * (100 – t0) / (U ^ 2 * t / R) = R * Sv * mw * (100 – t0) / (U ^ 2 * t), whence we express: mw = U2 * t * ηech / (R * Sv * (100 – t0)).
Const: St. – beats. heat capacity of water (Sv = 4.18 * 10 ^ 3 J / (kg * K).
Data: U – operating voltage (U = 220 V); t – the duration of the kettle (t = 10 min; in the SI system t = 600 s); ηech – the efficiency of the electric kettle (ηech = 95% = 0.95); R – heater resistance (R = 22 Ohm); t0 – early. temperature (t0 = 10 ºС).
Calculation: mv = U ^ 2 * t * ηech / (R * Sv * (100 – t0)) = 220 ^ 2 * 600 * 0.95 / (22 * 4.18 * 10 ^ 3 * (100 – 10) ) ≈ 3.3 (3) kg.
Answer: Bring 3.3 (3) kg of water to a boil.
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