0.2 kg of air is heated in a constant volume process from 20 to 100(degrees celsius). The specific heat at constant volume is 0.7186J/ Kg K.

Question

0.2 kg of air is heated in a constant volume process from 20 to 100(degrees celsius). The specific heat at constant volume is 0.7186J/ Kg K. Calculate the change in entropy of this process.

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Sigridomena 4 years 2021-07-29T09:20:25+00:00 1 Answers 15 views 0

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  1. Latifah
    0
    2021-07-29T09:21:56+00:00
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    Answer:

    \delta \ S =0.034679 \ J/K

    Explanation:

    The formula for calculating the change in  entropy of a process can be expressed as:

    \delta \ S = \int\limits \ \frac{d \theta }{T}

    \delta \ S = mc \int\limits^{T_2}_{T_1} \ \frac {d T}{T}}

    \delta \ S = mc \ In (T) ^{T_2}_{T_1}

    \delta \ S = mc \ [In \ (T_2) - In \ ({T_1})]

    \delta \ S = mc \ In \ (\frac{T_2}{T_1})

    Given that:

    mass m = 0.20 kg

    specific heat constant c  =  0.7186J/ Kg K

    T_2 = 100° C = ( 100 + 273.15) = 373.15 K

    T_1 = 20° C = ( 20  + 273.15) = 293.15 K

    Replacing our values into above equation; we have :

    \delta \ S = (0.20)(0.7186) \ In \ (\frac{373.15}{293.15})

    \delta \ S =0.034679 \ J/K

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