2 kg were heated in an aluminum pan weighing 700 g. water by 20 C. How much energy was spent on heating.

Given:

m1 = 700 grams = 0.7 kilograms – the mass of the aluminum pan;

m2 = 2 kilograms – the mass of water;

dt = 20 ° Celsius – the temperature at which the water was heated in an aluminum pan;

c1 = 900 J / (kg * C) – specific heat capacity of aluminum;

c2 = 4200 J / (kg * C) – specific heat capacity of water.

It is required to determine Q (Joule) – how much heat must be spent on heating water.

Since, according to the condition of the problem, the water is in an aluminum pan, both the water and the pan will be heated, that is:

Q = Qaluminum + Qwater;

Q = c1 * m1 * dt + c2 * m2 * dt;

Q = dt * (c1 * m1 + c2 * m2);

Q = 20 * (900 * 0.7 + 4200 * 2) = 20 * (630 + 8400) = 20 * 9030 = 180600 Joules (180.6 kJ).

Answer: it is necessary to expend energy equal to 180.6 kJ for heating.