What amount of heat did an aluminum saucepan weighing 200 g and the 1.5 liter water in it receive

What amount of heat did an aluminum saucepan weighing 200 g and the 1.5 liter water in it receive when heated from 20 C to 100 C?

ma = 200 g = 0.2 kg.

t1 = 20 ° C.

t2 = 100 ° C.

Vv = 1.5 l = 1.5 * 10 ^ -3 m ^ 3.

ρw = 1000 kg / m ^ 3.

Cw = 4200 J / kg * ° C.

Ca = 920 J / kg * ° C.

Q -?

The amount of heat Q will be the sum: Q = Q1 + Q2.

Where Q1 is the amount of heat that goes to heat the pan, Q2 is the amount of heat that goes to heat the water.

Q1 = Cа * ma * (t2 – t1).

Q1 = 920 J / kg * ° C * 0.2 kg * (100 ° C – 20 ° C) = 14720 J.

Q2 = Cw * mw * (t2 – t1).

We find the mass of water m by the formula: mw = Vw * ρw, where Vw is the volume of water, ρw is the density of water.

Q2 = Cw * Vw * ρw * (t2 – t1).

Q2 = 4200 J / kg * ° C * 1.5 * 10 ^ -3 m ^ 3 * 1000 kg / m ^ 3 * (100 ° C – 20 ° C) = 504000 J.

Q = 14720 J + 504000 J = 518720 J.

Answer: to heat the pot and water, they need to be told Q = 518720 J of thermal energy.