Determine the amount of heat that is needed in order to heat 4 kg of water from 20 degrees

Determine the amount of heat that is needed in order to heat 4 kg of water from 20 degrees to a boil in an aluminum pan weighing 500 g.

Given: m1 (weight of the aluminum pan) = 500 g (0.5 kg); m2 (mass of water) = 4 kg; t0 (initial temperature of water, pot) = 20 ºС.

Constants: C1 (specific heat capacity of aluminum) = 920 J / (kg * ºС); C2 (specific heat capacity of water) = 4200 J / (kg * ºС); tк (vaporization) = 100 ºС; L (specific heat of vaporization of water) = 2.3 * 10 ^ 6 J / kg.

Let’s compose the formula: Q = C1 * m1 * (tк – tн) + C2 * m2 * (tк – tн) + L * m2.

Calculation: Q = 920 * 0.5 * (100 – 20) + 4200 * 4 * (100 – 20) + 2.3 * 10 ^ 6 * 4 = 10580800 J ≈ 10.6 MJ.