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ΔT = 30°C - 29°C = 1 °C
Volumen de agua: 25m * 10 m * 1,8 m = 450 m^3
Ce = 4180 J/Kg*K = 4180J/Kg°C
densidad del agua, d = 1000 kg / m^3
b) Fórmula: Q = m * Ce * ΔT
c) Cálculos:
Q = 450m^3 * 1000 kg / m^3 * 4180 J/Kg°C * 1°C = 1.881.000.000 J = 1.881.000 KJ.
Respuesta: 1.881.000 KJ