Cold well water with a mass of 100 liters was poured into an iron shower tank, the mass of which was 60 kg.

Cold well water with a mass of 100 liters was poured into an iron shower tank, the mass of which was 60 kg. As a result of heating by solar radiation, the water temperature increased from 5 degrees to 35 degrees how much heat the tank and water received.

Given:
m1 = 60 kg – the mass of the iron tank
V2 = 100 l = 0.1 m ^ 3 – volume of water
t1 = 5 C
t2 = 35 C
C1 = 460 J / kg * С – specific heat of iron
С2 = 4200 J / kg * С – specific heat of water
p = 1000 kg / m ^ 3 – water density
Q -?
Q1 = C1 * m1 * (t2 – t1) – the amount of heat required to heat the iron tank
Q2 = C2 * m2 * (t2 – t1) – the amount of heat required to heat water
m2 = V2 * p
Q1 = 460 * 60 * (35-5) = 828000 J
m2 = 0.1 * 1000 = 100 kg
Q2 = 4200 * 100 * (35-5) = 12600000 J
Q = 828000 + 12600000 = 13428000 J = 13428 kJ