How much heat is required to heat 6 liters of water in an aluminum pan of m = 700 g from 20 degrees to boiling?

ma = 700 g = 0.7 kg.

t1 = 20 ° C.

t2 = 100 ° C.

Vv = 6 l = 6 * 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 * 0.7 kg * (100 ° C – 20 ° C) = 51520 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 * 6 * 10 ^ -3 m ^ 3 * 1000 kg / m ^ 3 * (100 ° C – 20 ° C) = 2016000 J.

Q = 51520 J + 2016000 J = 2067520 J.

Answer: for heating you need Q = 2067520 J of thermal energy.