The specific heat of lead is 0.129 J/gC. Find the amount of heat released when 2.4 moles of lead are cooled from 37.2C to 22.5C.

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The specific heat of lead is 0.129 J/gC. Find the amount of heat released when 2.4 moles of lead are cooled from 37.2C to 22.5C.

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Lệ Thu 4 years 2021-08-12T16:14:31+00:00 1 Answers 10 views 0

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  1. Nho
    0
    2021-08-12T16:15:32+00:00
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    Answer:

    -943J or 943J released

    Explanation:

    m=2.4 mol x 207.g/ mol=497.3g

    c=.129 J/gC

    T= T f-Ti=(22.5-37.2)=-14.7C

    Q=?

    Q=(49.3g)(.129 J/g C) (-14.7 C)

    = -943 J or 943 J released

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