To heat 0.5 kg of water on an alcohol lamp from 20 ° C to 100 ° C and transfer 30 g to steam

To heat 0.5 kg of water on an alcohol lamp from 20 ° C to 100 ° C and transfer 30 g to steam, what amount of alcohol was burned? 30% was involuntarily spent.

0.7Q1 = Q2 + Q3 (cost accounting: 70% = 100% – 30% = 0.7).

Q1 (alcohol) = q * m1, where q = 2.7 * 10 ^ 7 J / kg, m2 is the desired mass of alcohol spent.

Q2 (water heating) = С * m2 * (tк – tн), where С = 4200 J / (kg * К), m2 = 0.5 kg, tк = 100 ºС, tн = 20 ºС.

Q3 (vaporization) = L * m3, where L = 2.3 * 10 ^ 6 J / kg, m = 30 g or 0.03 kg.

0.7q * m1 = С * m2 * (tк – tн) + L * m3.

m1 = (С * m2 * (tк – tн) + L * m3) / (0.7 * q) = (4200 * 0.5 * (100 – 20) + 2.3 * 10 ^ 6 * 0.03 ) / (0.7 * 2.7 * 10 ^ 7) = 0.0125 kg = 12.5 g.