# At what maximum internal resistance is a source with an emf of 70 V able to boil 5 kg

June 30, 2021 | education

| **At what maximum internal resistance is a source with an emf of 70 V able to boil 5 kg of water taken at 313 K in 1 hour?**

Data: ε (source EMF) = 70 V; m (mass of boiled water) = 5 kg; t1 (initial temperature) = 313 K (40 ºС); t (duration of source operation) = 1 h = 3600 s; the external resistance of the source is taken to be zero (R = 0 Ohm).

Constants: t2 (boiling point of water) = 100 ºС; Sv (specific heat of water) = 4200 J / (kg * K).

Solution: Sv * m * (t2 – t1) = I ^ 2 * r * t = (ε / (R + r)) ^ 2 * r * t = ε ^ 2 * t / r, whence r = ε ^ 2 * t / Sv * m * (t2 – t1) = 70 ^ 2 * 3600 / (4200 * 5 * (100 – 40)) = 14 ohms.

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