In an aluminum saucepan weighing 2.2 kg. heated to a boil 2 liters. water, 100 degrees Celsius

In an aluminum saucepan weighing 2.2 kg. heated to a boil 2 liters. water, 100 degrees Celsius, the initial temperature of which is 0 degrees Celsius. How much heat was required?

mh = 2.2 kg.

Ca = 920 J / kg * ° C.

Vw = 2 l = 0.002 m3.

Cw = 4200 J / kg * ° C.

ρw = 1000 kg / m3.

t1 = 0 ° C.

t2 = 100 ° C.

Q -?

When heated, the heat energy goes to heat the aluminum kettle and the water in it, so the required amount of heat will be the sum: Q = Q1 + Q2.

The amount of heat Q1 that goes to heat the kettle is expressed by the formula: Q1 = Ca * mh * (t2 – t1).

Q1 = 920 J / kg * ° C * 2.2 kg * (100 ° C – 0 ° C) = 202400 J.

The amount of heat Q2, which goes to heating water, is expressed by the formula: Q2 = Cw * mw * (t2 – t1).

We express the mass of heated water mw by the formula: mw = Vw * ρw.

Q2 = Cw * Vw * ρw * (t2 – t1).

Q2 = 4200 J / kg * ° C * 0.002 m3 * 1000 kg / m3 * (100 ° C – 0 ° C) = 840,000 J.

Q = 202400 J + 840,000 J = 1042500 J.

Answer: for heating, you need Q = 1042400 J of thermal energy.