To what temperature will 2 kg of water be heated at a temperature of 14 grv if only 60% of the heat is used for heating

To what temperature will 2 kg of water be heated at a temperature of 14 grv if only 60% of the heat is used for heating, you are released during the combustion of alcohol weighing 20 kg.

t1 = 14 ° C.
C = 4200 J / kg * ° C.
mw = 2 kg.
mс = 20 kg.
q = 27 * 106 J / kg.
Efficiency = 60%.
t2 -?
Let us express the amount of heat Qс, which is released during the combustion of a sleeping bag, according to the formula: Qс = q * mс, where q is the specific heat of combustion of alcohol, mс is the mass of alcohol that burned out.
Let us express the amount of heat Qw, which is necessary for heating water, according to the formula: Qw = C * mw * (t2 – t1).
Since only efficiency = 60% of thermal energy during combustion goes to heating water, then we write it mathematically: Qw = Qc * efficiency / 100%.
C * mw * (t2 – t1) = q * ms * efficiency / 100%.
C * mw * t2 – C * mw * t1 = q * mc * efficiency / 100%.
t2 = (q * ms * efficiency / 100% + C * mv * t1) / C * mv.
t2 = (27 * 106 J / kg * 20 kg * 60% / 100% + 4200 J / kg * ° C * 2 kg * 14 ° C) / 4200 J / kg * ° C * 2 kg = 38585.5 ° C.
Answer: water can be heated to a temperature of t2 = 38585.5 ° C.