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2 August, 07:19

5. How much heat (in kJ) is given out when 85.0 g of lead cools from 200.0

°C to 10.0°C? (Cp of lead = 0.129 J/g °C)

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Answers (1)
  1. 2 August, 07:40
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    2.08335 kj

    Explanation:

    Given dа ta:

    Mass of lead = 85 g

    Initial temperature = 200°C

    Final temperature = 10°C

    Heat lost = ?

    Cp of lead = 0.129 j/g.°C

    Solution:

    Specific heat capacity: Cp

    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 = 10°C - 200°C

    ΔT = - 190°C

    Now we will put the values:

    Q = 85 g * 0.129 j/g.°C * - 190°C

    Q = 2083.35 j

    Joule to kilo joule:

    2083.35/1000 = 2.08335 kj
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