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A 15.0-kg block of ice at 0.0°C melts to liquid water at 0.0°C inside a large room at 20.0°C. Treat the ice and the room as an isolated system, and assume that the room is large enough for its temperature change to be ignored. Calculate the net entropy change of the system during this process.

My attempt:

Due to the finite temperature difference between the ice and the room, the melting of the ice is irreversible.

Since entropy is a function only of the state of a system, we can also compute entropy changes in irreversible processes for which the usual equations are not applicable. We must invent a path connecting the given initial and final states that does consist entirely of reversible equilibrium processes and compute the total entropy change for that path. But how do I go about doing this in this example?

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    $\begingroup$ You bring the block of ice into contact with a constant temperature reservoir that is at a temperature slightly above 0 C. In this case, all the heat transfer would take place reversibly. $\endgroup$ – Chet Miller Jul 9 at 16:10
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We don’t solve homework and exercise problems on this site. But here’s some guidance.

Do you know the equation for a differential change in entropy, $dS$? If you do here’s a hint:

Heat transfer out of the room and into the ice both occur isothermally. The total charge in entropy is

$$\Delta S_{tot}=\Delta S_{ice}+\Delta S_{room}$$

Hope this helps

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