Electric stove heaters 1.2 liters of water from 10’C to 100’C. At the same time, 3% of it turned into steam.

Electric stove heaters 1.2 liters of water from 10’C to 100’C. At the same time, 3% of it turned into steam. How long did the heating take if the power of the tile was 800 W and its efficiency was 65%?

Net power of the hotplate P:
P = kP₀,
where k is the efficiency, P₀ is the power of the hotplate.
The amount of heat Q required to heat and evaporate water
Q = cm (T-T₀) + r * 0.03m,
where c is the heat capacity of water,
m is the mass of water (the mass of one liter is 1 kg; 1, 2 l – 1.2 kg)
0,03m – steam mass,
T₀ and T – initial and final water temperature,
r is the specific heat of vaporization.
Let the heating and evaporation time be t. Then the amount of heat taken from the tile during time t is equal to the amount of heat for heating and evaporation:
Pt = Q;
kP₀t = cm (T-T₀) + 0.03rm;
t = (cm (T-T₀) + 0.03rm) / kP₀;
t = (4200 J / kg * deg * 1.2 kg * (100 ° C – 10 ° C) + 0.03 * 2.3 * 10 ^ 6 J / kg * 1.2 kg) / (0.65 * 800 W)
= 1031 s.
Answer: It will take 1031 s (17 min 11 s).