On a stove, 4 kg of water was heated to a boil, taken at t = 20 degrees and completely

On a stove, 4 kg of water was heated to a boil, taken at t = 20 degrees and completely evaporated. Determine how much kerosene was required for this?

Let’s write the heat balance equation:
Q1 = Q2 + Q3,
Where
Q1 = m * q – the amount of heat that is obtained during the combustion of kerosene;
Here
m is the required mass,
q = 44 * 10 ^ 6 J / kg – specific heat of combustion of kerosene;
Q2 = M * c * (t2 – t1) – the amount of heat for heating water from t1 = 20 ° С to t2 = 100 ° С;
M = 4kg – mass of water,
c = 4200J / (kg * deg) – specific heat capacity of water;
Q3 = M * r – the amount of heat for the evaporation of water,
r = 2.26 * 10 ^ 6 – specific heat of vaporization;
m = M * [r + c * (t2 – t1)] / q;
m = 4 * (2.26 * 10 ^ 6 + 336000) / 44 * 10 ^ 6 = 0.236kg = 236g.