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    Calorimetry Pre Lab

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    Assume you use calorimetry to calculate the specific heat capacity of a 125.24 g piece of unknown metal. You intially heat the metal to 100.0 °C in boiling water. You then drop the chunk of metal into a calorimeter containing 47.22 g of water at 19.7 °C. After closing and stiring the calorimeter thoroughly, the metal and water both come to equilibrium at a temperature of 27.6 °C.
    Part A
    What is the temperature change of the water?
    100.0 °C
    19.7 °C
    7.9 °C

    Part B
    What is the temperature change of the metal?
    -72.4 °C
    125.2 °C
    -27.6 °C

    Part C
    How much heat was gained by the water? (calculate the qwater)
    33.1 J
    1561 J
    418.4 J

    Part D
    Knowing the above, what must qmetal?
    -33.1 J
    -1561 J
    -125240 J

    Part E
    Then what must the the Specific Heat of the metal be?
    0.4184 J/g°C
    25.00 J/g°C
    0.1721 J/g°C

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