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28 April, 07:45

The specific heat capacity of lead is 0.13 J/g-°C. How much heat (in J) is required to raise the temperature of of 91 g of lead from 22°C to 37°C?

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  1. 28 April, 07:59
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    Q = 177J

    Explanation:

    Specific heat capacity of lead=0.13J/gc

    Q=MCΔT

    ΔT=T2-T1, where T1=22degrees Celsius and T2=37degree Celsius.

    ΔT=37 - 22 = 15

    Q = Change in energy

    M = mass of substance

    C = Specific heat capacity

    Q = (91g) * (0.13J/gc) * (15c) = 177.45J

    Approximately, Q = 177J
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