Consider a room air conditioner using a Carnot cycle at maximum theoretical efficiency and operating between the temperatures of \(18^{\circ} \mathrm{C}\) (indoors) and \(35^{\circ} \mathrm{C}\) (outdoors). For each 1.00 J of heat flowing out of the room into the air conditioner: a) How much heat flows out of the air conditioner to the outdoors? b) By approximately how much does the entropy of the room decrease? c) By approximately how much does the entropy of the outdoor air increase?

To work with temperature values in this case, it's necessary to convert them to Kelvin. To do that, add 273.15 to each Celsius value to find their equivalent Kelvin temperatures. \(T_{room} = 18^{\circ} \mathrm{C} + 273.15 = 291.15\,\mathrm{K}\) \(T_{outdoor} = 35^{\circ} \mathrm{C} + 273.15 = 308.15\,\mathrm{K}\)

The efficiency of a Carnot cycle is given by: $$ \eta = 1 - \frac{T_{low}}{T_{high}} $$ Here, \(T_{low}\) is the lower temperature (which is the room temperature) and \(T_{high}\) is the higher temperature (the outdoor temperature). Plug in the given values to find the efficiency: $$ \eta = 1 - \frac{291.15\,\mathrm{K}}{308.15\,\mathrm{K}} = 1 - 0.9447 \approx 0.0553 $$

Let \(W\) be the work done by the air conditioner, and \(Q_{room}\) the heat removed from the room (which is given), then \(Q_{room} = 1.00\,\mathrm{J}\). Since we are calculating at maximum theoretical efficiency, we can use the formula: $$ \eta = \frac{W}{Q_{room}} $$ Rearranging the equation to get \(W\): $$ W = \eta \times Q_{room} = 0.0553 \times1.00\,\mathrm{J} = 0.0553\,\mathrm{J} $$ Now, conservation of energy states that the heat transferred to the outdoor (\(Q_{outdoor}\)) plus the work done (\(W\)) should equal the heat removed from the room (\(Q_{room}\)). Thus: $$ Q_{outdoor} = Q_{room} - W = 1.00\,\mathrm{J} - 0.0553\,\mathrm{J} \approx 0.9447\,\mathrm{J} $$ So, the heat flowing out of the air conditioner to the outdoors is approximately 0.9447 J.

We can use the formula for entropy change (\(\Delta S\)), which is $$ \Delta S = \frac{Q}{T} $$ For the room, the entropy change is: $$ \Delta S_{room} = \frac{Q_{room}}{T_{room}} = \frac{1.00\,\mathrm{J}}{291.15\,\mathrm{K}} \approx 0.00344\,\mathrm{J/K} $$ Now, for the outdoor entropy change: $$ \Delta S_{outdoor} = \frac{Q_{outdoor}}{T_{outdoor}} = \frac{0.9447\,\mathrm{J}}{308.15\,\mathrm{K}} \approx 0.00307\,\mathrm{J/K} $$ So, the entropy of the room decreases by approximately 0.00344 J/K while the entropy of the outdoor air increases by approximately 0.00307 J/K.