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24 May, 10:19

Aluminum has a specific heat of 0.902 j/gC. How much heat is lost when a piece of aluminum with a mass of 23.984 g cools from a temperature of 415 degrees to 22 degrees?

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  1. 24 May, 10:45
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    Q = - 8501.99 j

    Explanation:

    Given dа ta:

    Specific heat of Al = 0.902 j/g.°C

    Heat lost = ?

    Mass of sample = 23.984 g

    Initial temperature = 415°C

    Final temperature = 22°C

    Solution:

    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

    ΔT = 22°C - 415°C

    ΔT = - 393°C

    Q = m. c. ΔT

    Q = 23.984 g * 0.902 j/g.°C * - 393°C

    Q = - 8501.99 j
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