What amount of heat did an aluminum pan weighing 200 g receive the water in it with a volume

What amount of heat did an aluminum pan weighing 200 g receive the water in it with a volume of 1.5 liters when it was heated from 20 degrees to 100 degrees?

The pot and water received a certain amount of heat from the heating source, as evidenced by their heating to 100 degrees Celsius.
The amount of heat received by the body when heated to a certain temperature is determined by the formula: Q = c * m * (t heat – t start), where Q is the amount of heat received when the body is heated, c is the specific heat capacity of its substance (in our case, c aluminum = 920 J / kg ° C, and c water = 4200 J / kg ° C), m is the body weight (in our case, water has a mass of 1.5 kg, since the ρ of water is 1000 kg / m³ or 1 kg / l), t heat is the temperature to which the body was heated, t start is the initial temperature at which the bodies began to be heated (in our case, the temperature difference has already been given (t heat – t start) = 100 ° С – 20 ° С = 80 ° C).
Substituting the values ​​we know, we find that
the amount of heat received by the pan will be – 920 J / kg ° C * 0.2 kg * 80 ° C = 14720 J or 14.72 kJ.
The amount of heat received by water will be –
4200 J / kg ° C * 1.5 kg * 80 ° C = 504000 J or 504 kJ.