How much water can be heated in an aluminum vessel weighing 1 kg from 20 degrees to boiling (1800 degrees)

How much water can be heated in an aluminum vessel weighing 1 kg from 20 degrees to boiling (1800 degrees), burning 100 g of gas in an installation with an efficiency of 40% (c.al = 880 J / kg, q = 46 MJ / kg)

mk = 1 kg.

t1 = 20 ° C.

t2 = 100 ° C.

Cw = 4200 J / kg * ° С.

Ca = 880 J / kg * ° С.

Efficiency = 40%.

qg = 46 MJ = 46 * 10 ^ 6 J / kg.

mg = 100 g = 0.1 kg.

mv -?

Efficiency = Qw * 100% / Qg.

Qw = Sv * mw * (t2 – t1) + Ca * mk * (t2 – t1).

Qg = qg * mg.

Efficiency = (Sv * mw * (t2 – t1) + Ca * mk * (t2 – t1)) * 100% / qg * mg.

Efficiency * qg * mg / 100% = Sv * mw * (t2 – t1) + Ca * mk * (t2 – t1).

mv = (efficiency * qg * mg / 100% – Ca * mk * (t2 – t1)) / Sv * (t2 – t1). =

mw = (40% * 46 * 10 ^ 6 J / kg * 0.1 kg / 100% – 880 J / kg * ° C * 1 kg * (100 ° C – 20 ° C)) / 4200 J / kg * ° С * (100 ° С – 20 ° С) = 5.27 kg.

Answer: you can heat mw = 5.27 kg of water.