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3 January, 05:21

A 52.0-g sample of an unknown metal at 99°C was placed in a constant-pressure calorimeter containing 75.0 g of water at 24.0°C. The final temperature of the system was found to be 28.4°C. Calculate the specific heat of the metal. (The heat capacity of the calorimeter is 14.4 J/°C.)

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  1. 3 January, 08:24
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    The specific heat of the metal is 0.39251 J/g°C

    Explanation:

    The heat transfered from the metal to the calorimeter and water (Qwater + Qcal), must be equal to Qmetal (but different in direction).

    Qmetal = - (Qwater + Qcal)

    For Q we use the following formule Q = m * Cp*ΔT

    with m = the mass in grams

    with Cp = the heat capacity in J / g°C

    with ΔT = the change in temperature = Final temperature T2 - initial temperature T1

    Qmetal = m (Metal) * Cp (metal) * ΔT (metal)

    with mass = 52 grams

    with Cp = To be determined

    with ΔT = 28.4 - 99 = - 70.6 °C

    QCal = Cp (cal) * ΔT (water)

    Qcal = 14.4*4.184 = 60.2496

    Qwater = m (water) * Cp (water) * ΔT (water)

    with mass = 75 g

    with Cp (water) = 4.184

    with ΔT (water) = 28.4 - 24 = 4.4

    Qwater = 1380.72

    Q (metal) = - (Qwater + Qcal)

    m (Metal) * Cp (metal) * ΔT (metal) = - 60.2496 - 1380.72

    Cp (metal) = - 1440.9696 / (52 * - 70.6)

    Cp (metal) = - 1440.9696 / - 3671.2

    Cp (metal) = 0.39251 J/g°C

    The specific heat of the metal is 0.39251 J/g°C
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