It took 20 minutes to heat a certain amount of water from 30 to 100 degrees on an electric stove.

It took 20 minutes to heat a certain amount of water from 30 to 100 degrees on an electric stove. How long does it take then for the complete evaporation of the water.

T1 = 303 K.
T2 = 373 K.
t1 = 20 min = 1200 s.
c = 4187 J / kg * K.
λ = 2260000 J / kg.
t2 -?
To heat water to the boiling point, you need Q1 of the amount of heat, which is determined by the formula: Q1 = c * m * (T2 -T1), where c is the specific heat capacity of water, m is the mass of water.
Let us determine the power N that the electric stove has: N = Q1 / t1, where Q1 is the amount of heat transferred by the stove, t1 is the time during which this heat is transferred to the water.
N = c * m * (T2 -T1) / t1.
During boiling, the heater has the same power and for boiling it transfers the amount of heat Q2 = N * t2, where t2 is the time during which all the water will boil away.
Q2 = λ * m, where λ is the specific heat of vaporization, m is the mass of water.
λ * m = c * m * (T2 -T1) * t2 / t1.
t2 = λ * t1 / c * (T2 -T1).
t2 = 2260000 J / kg * 1200 s / 4187 J / kg * K * (373 K -303 K) = 9253 s.
Answer: complete evaporation takes time t2 = 9253 s.