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25 June, 09:10

The temperature of a sample of lead increased by 22.9 °C when 265 J of heat was applied. What is the mass of the sample?

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Answers (1)
  1. 25 June, 09:49
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    90.41 g

    Explanation:

    Data provided in the question:

    Increase in temperature of lead sample, ΔT = 22.9 °C

    Heat supplied to the sample, Q = 265 J

    Now,

    we know that

    Q = mCΔT

    here,

    m = mass of the sample

    C = Specific heat of lead = 0.128 J/g.°C

    Thus,

    265 = m * 0.128 * 22.9

    or

    265 = m * 2.9312

    or

    m = 90.41 g
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