This problem is describing a system in which 100 kJ of energy is transferred from a cold reservior at 600 K to a hot one at 1050 K. This situation is opposed to the Clausius statement of the second law of thermodynamics, because it is widely known that heat is transferred from hot to cold systems.
However, we can calculate the total entropy change by using the following formula:
[tex]\Delta S =\Delta S _{Low}+\Delta S _{High}[/tex]
Which is broken down as follows:
[tex]\Delta S =\frac{-Q}{T_{Low}} + \frac{Q}{T_{High}}\\\\\\\Delta S =Q(\frac{-1}{T_{Low}} + \frac{1}{T_{High}})[/tex]
Since the heat is assumed to be equal for the both of them. Then, we plug in the given heat and temperatures to obtain:
[tex]\Delta S =100kJ(\frac{-1}{600K} + \frac{1}{1050K})\\\\\Delta S =-0.0714\frac{kJ}{K}[/tex]
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