2 liters of water were poured into a saucepan at a temperature of 15 degrees. After boiling, there was 200g less water in the pan than at the beginning of heating. How much heat did the water get in the pot?
To calculate the amount of thermal energy received by water, we use the formula: Q = Q1 (evaporation) + Q2 (heating) = L * mi + Sv * mn * (tboil – t) = L * mi + Sv * ρw * V * (tboil – t).
Data: L – beats heat of vaporization (L = 2.3 * 10 ^ 6 J / kg); mi is the mass of evaporated water (mi = 200 g = 0.2 kg); St. – beats. heat capacity of water (Sv = 4.2 * 10 ^ 3 J / (kg * ºС)); ρw is the density of the poured water (ρw = 10 ^ 3 kg / m3); V is the initial volume of water (V = 2 l = 2 * 10 ^ -3 m3); tboil – boiling point (tboil = 100 ºС); t – initial temperature (t = 15 ºС).
Calculation: Q = 2.3 * 10 ^ 6 * 0.2 + 4.2 * 10 ^ 3 * 10 ^ 3 * 2 * 10 ^ -3 * (100 – 15) = 1174000 J = 1.174 MJ.
Answer: The water in the pot should have received 1.174 MJ of thermal energy.
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