An aluminum kettle weighing 0.8 kg contains 3 liters of water at 20 ° C. Determine the amount of heat that needs

An aluminum kettle weighing 0.8 kg contains 3 liters of water at 20 ° C. Determine the amount of heat that needs to be spent on heating the kettle with water to a boil.

mh = 0.8 kg.
Vw = 3 l = 0.003 m3.
ρw = 1000 kg / m3.
Cw = 4200 J / kg * ° C.
t1 = 20 ° C.
t2 = 100 ° C.
Ca = 920 J / kg * ° C.
Q -?
When heated, the amount of heat Q goes to heating the aluminum kettle Qh and water Qw, therefore: Q = Qh + Qw.
The amount of heat Q required to heat a substance is determined by the formula: Q = C * m * (t2 – t1), where C is the specific heat capacity of the substance, m is the mass of the heated substance, t2, t1 are the final and initial temperatures during heating.
Qh = Ca * mh * (t2 – t1).
Qh = 920 J / kg * ° C * 0.8 kg * (100 ° C – 20 ° C) = 58880 J.
Qw = Cw * mw * (t2 – t1).
We express the mass of water mw by the formula: mw = Vw * ρw.
Qw = Cw * Vw * ρ * (t2 – t1).
Qw = 4200 J / kg * ° C * 0.003 m3 * 1000 kg / m3 * (100 ° C – 20 ° C) = 1008000 J.
Q = 58880 J + 1008000 J = 1066880 J.
Answer: to heat a kettle with water, you need to spend Q = 1066880 J of thermal energy.