In an aluminum pan, weighing 800 g, water is heated, with a volume of 5 liters from 100 C to boiling.

In an aluminum pan, weighing 800 g, water is heated, with a volume of 5 liters from 100 C to boiling. How much heat will be used to heat the pot and water? s (water) = 4200 J / kg K, s (aluminum) = 920 J / kg K p = 1000 kg / m3, t = 1000 C.

Vv = 5 l = 5 * 10 ^ -3 m ^ 3.

Ca = 920 J / kg * ° C.

ρv = 1000 kg / m ^ 3.

Sv = 4200 J / kg * ° C.

t1 = 10 ° C.

t2 = 100 ° C.

Q -?

The amount of heat Q will be the sum of Q1 and 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.

Q = Q1 + Q2,

Q1 is determined by the formula: Q1 = Ca * mk * (t2 – t1), where Ca is the specific heat capacity of aluminum, mk is the mass of the pan, t2, t1 are the final and initial temperature of the pan.

Q2 is determined by the formula: Q2 = Cw * mw * (t2 – t1), where Cw is the specific heat capacity of water, mk is the mass of water, t2, t1 are the final and initial water temperatures.

mw = Vw * ρw.

Q2 = Cw * Vw * ρw * (t2 – t1).

Q = Ca * mk * (t2 – t1) + Cw * Vw * ρw * (t2 – t1) = (Ca * mk + Cw * Vw * ρw) * (t2 – t1).

Q = (920 J / kg * ° C * 0.8 kg + 4200 J / kg * ° C * 5 * 10 ^ -3 m ^ 3 * 1000 kg / m ^ 3) * (100 ° C – 10 ° C ) = 1956240 J.

Answer: to heat the pot and water, you need Q = 1956240 J.