The kettle is plugged into a 220 V network. What is the efficiency of the kettle if the current

| September 17, 2021 | education

The kettle is plugged into a 220 V network. What is the efficiency of the kettle if the current in its coil is 7 A and it can be heated from 20 ° C to boiling 2.2 kg of water in 10 minutes?

Kettle efficiency:
η = Q / A, where Q is the amount of heat that goes to heat the water, A is the work of the electric current.
Q = c * m * ∆T, where c is the specific heat capacity of water (c = 4183 J / (kg * K)), m is the mass of water (m = 2.2 kg), ∆T is the difference in water temperatures (∆ T = 100 – 20 = 80 ºС).
A = I * U * t, where I is the current strength (I = 7 A), U is the voltage in the network (U = 220 V), t is the operating time of the kettle (t = 10 min = 10 * 60 s = 600 s ).
η = Q / A = s * m * ∆T / (I * U * t) = 4183 * 2.2 * 80 / (7 * 220 * 600) = 0.797 = 79.7%.
Answer: The efficiency of the kettle is 79.7%.



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