How much alcohol will you need to warm a kilogram of water from 10 to 20 degrees?

mw = 1 kg
sv = 4200 J / kg * C – specific heat of water
t1 = 10 C
t2 = 20 C
qс = 2.7 * 10 ^ 7 J / kg – specific heat of combustion of alcohol
mc -?
we use the heat balance equation Qc = Qw
Qc = q * mc
Qw = C * mw * (t2 – t1)
q * mc = C * mw * (t2-t1)
mc = C * mw * (t2-t1) / q
mc = 4200 * 1 * (20-10) / 2.7 * 10 ^ 7 = 0.0016 kg = 1.6 g