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11 October, 01:00

15.5 g of an unknown metal at 165.0°C is dropped into 150.0mL of H2O at 23.0°C in a coffee cup calorimeter. The metal and H2O reached thermal equilibrium at 30.0°C. Calculate the specific heat capacity of the unknown metal. Specific heat of water is 4.184 J/g°C.

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  1. 11 October, 01:17
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    Specific heat capacity of metal is 2.09 j/g.°C.

    Explanation:

    Specific heat capacity:

    It is the amount of heat required to raise the temperature of one gram of substance by one degree.

    Formula:

    Q = m. c. ΔT

    Q = amount of heat absorbed or released

    m = mass of given substance

    c = specific heat capacity of substance

    ΔT = change in temperature

    Given dа ta:

    Mass of metal = 15.5 g

    Initial temperature = 165.0°C

    Initial temperature of water = 23.0°C

    Final temperature = 30.0°C

    Specific heat capacity of metal = ?

    Specific heat capacity of water = 4.184 J/g°C

    Volume of water = 150.0 mL or 150.0 g

    Solution:

    Formula:

    - Qm = + Qw

    Now we will put the values in formula.

    -15.5 g * c * [ 30.0°C - 165.0°C] = 150 g * 4.184 J/g°C * [ 30.0°C - 23.0°C]

    15.5 g * c * 135°C = 4393.2 j

    2092.5 g.°C * c = 4393.2 j

    c = 4393.2 j/2092.5 g.°C

    c = 2.09 j/g.°C
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