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22 December, 08:09

Calculate the specific heat of a substance if a 35g sample absorbs 48 j as the temperature is raised from 293 k to 313 k

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  1. 22 December, 09:03
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    c = 0.07 j/g. k

    Explanation:

    Given dа ta:

    Mass of sample = 35 g

    Heat absorbed = 48 j

    Initial temperature = 293 K

    Final temperature = 313 K

    Specific heat of substance = ?

    Solution:

    Specific heat capacity:

    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 = Final temperature - initial temperature

    ΔT = 313 k - 293 K

    ΔT = 20 k

    Now we will put the values in formula.

    48 j = 35 g * c * 20 k

    48 j = 700 g. k * c

    c = 48 j/700 g. k

    c = 0.07 j/g. k
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