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17 October, 15:59

Una muestra de 500g de agua se calienta desde 10°C hasta 80°C. Calcula la cantidad de calor absorbido por líquido si su calor específico es 4186J/kg * K

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  1. 17 October, 17:04
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    Q = 142.324kJ

    Explanation:

    dа ta:

    M = 500g = 0.5kg

    T1 = 10°C = (10 + 273.15) K = 285.15K

    T2 = 80°C = (80 + 273.15) K = 353.15K

    Q = ?

    C = 4186J/kg. K

    Q = mc (T2 - T1)

    Q = 0.5 * 4186 * (353.15 - 285.15)

    Q = 0.5 * 4186 * 68

    Q = 142324J

    Q = 142.324kJ.
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