An electric heater brings 2.5 kg of water to a boil in 15 minutes, the initial temperature of which is 10 ° C.

An electric heater brings 2.5 kg of water to a boil in 15 minutes, the initial temperature of which is 10 ° C. The current in the heater is 7 A, the efficiency of the heater is 60%. What is the voltage in the electrical network? The specific heat capacity of water is 4200 J / (kg ° C)

T = 15 min = 900 s.
m = 2.5 kg.
t1 = 10 ° C.
t2 = 100 ° C.
I = 7 A.
η = 60%.
c = 4200 J / (kg * ° C).
U -?
Heater efficiency η is determined by the formula: η = Qw * 100% / Qn. Where Qw is the amount of heat that is communicated to the water for heating, Qn is the amount of heat that the heater emits.
The amount of heat that goes into heating water is determined by the formula: Qw = c * m * (t2 – t1).
The amount of heat that the heater emits is determined by the Joule-Lenz law: Qн = U * I * T, where U is the voltage, I is the current in the heater, T is the time of passage of the electric current.
η = c * m * (t2 – t1) * 100% / U * I * T.
U = c * m * (t2 – t1) * 100% / η * I * T.
U = 4200 J / (kg * ° C) * 2.5 kg * (100 ° C – 10 ° C) * 100% / 60% * 7 A * 900 s = 250 V.
Answer: the voltage in the electrical network is U = 250 V.