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22 September, 20:49

A burner on an electric range has a heat capacity of 345 J/c. what value of q, in kJ, is produced when the burner cools from a temperature of 467 C to room temperature of 23 C?

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  1. 22 September, 23:41
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    -153.18 kj

    Explanation:

    Given dа ta:

    Heat capacity of burner = 345 j/°C

    Initial temperature = 467°C

    Final temperature = 23°C

    Heat produced = ?

    Solution:

    Formula:

    q = cΔT

    q = heat absorbed or released

    c = heat capacity

    ΔT = change in temperature

    ΔT = 23°C - 467°C

    ΔT = - 444°C

    q = cΔT

    q = 345 j/°C * - 444°C

    q = - 153180 J

    joule to kj:

    q = - 153180 J * 1 kj / 1000 j

    q = - 153.18 kj
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