m1 = 800 grams = 0.8 kilograms – the mass of the aluminum pan;
V = 5 liters = 0.005 m3 (cubic meters) – the volume of water in the pan;
ro = 1000 kg / m3 (kilogram per cubic meter) – density of water (clean);
dt = 10 ° Celsius – the temperature at which the water was heated in an aluminum pan;
c1 = 900 J / (kg * C) – specific heat 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, where m2 is the mass of water;
Q = dt * (c1 * m1 + c2 * ro * V);
Q = 10 * (900 * 0.8 + 4200 * 1000 * 0.005) = 20 * (720 + 21000) = 20 * 21720 = 434400 Joules (434.4 kJ).
Answer: it is necessary to expend energy equal to 434.4 kJ for heating.
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