Question

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?

Short answer

Answer: The heat flowing out of the air conditioner to the outdoors is approximately 0.9447 J. The entropy of the room decreases by approximately 0.00344 J/K, and the entropy of the outdoor air increases by approximately 0.00307 J/K.

Step-by-step solution

Get the temperatures in Kelvin

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. \(T_{room} = 18^{\circ} \mathrm{C} + 273.15 = 291.15\,\mathrm{K}\) \(T_{outdoor} = 35^{\circ} \mathrm{C} + 273.15 = 308.15\,\mathrm{K}\)

Determine the efficiency for the Carnot cycle

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: $$ \eta = 1 - \frac{291.15\,\mathrm{K}}{308.15\,\mathrm{K}} = 1 - 0.9447 \approx 0.0553 $$

Calculate the heat flowing out of the air conditioner

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.

Calculate the entropy decrease in the room and increase in the outdoor air

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.

Key concepts

Thermodynamic Efficiency

Thermodynamic efficiency is a critical concept in understanding how a system like a Carnot cycle operates. In simple terms, efficiency in thermodynamics gauges how well a heat engine converts heat into work. For the idealized Carnot engine, the efficiency depends solely on the temperatures of the hot and cold reservoirs. The efficiency \( \eta \) is given by the formula:
\[\eta = 1 - \frac{T_{\text{low}}}{T_{\text{high}}}\]
where \( T_{\text{low}} \) and \( T_{\text{high}} \) are the absolute temperatures of the cold and hot reservoirs, respectively, expressed in Kelvin.

In the given exercise, it illustrates that even the most theoretically perfect engine, the Carnot cycle, cannot achieve 100% efficiency due to these temperature constraints. This is a key principle of the Second Law of Thermodynamics, emphasizing why real systems always consume more energy than what they convert. Understanding efficiency is essential for improving energy systems and reducing waste.

Entropy Change

Entropy is often described as a measure of disorder or randomness in a system, but it's important to think of it in terms of energy dispersal at a particular temperature. In thermodynamics, the change in entropy provides insight into the inefficiencies of energy transformations.

For a system undergoing a heat transfer \( Q \) at an absolute temperature \( T \), the entropy change \( \Delta S \) can be calculated using
\[\Delta S = \frac{Q}{T}\]This formula offers a way to quantify how energy transitions affect the system's entropy.

• **Room Entropy Change**: When 1.00 J of heat leaves the room, the decrease in entropy is approximately 0.00344 J/K, according to the exercise calculations.
• **Outdoor Air Entropy Change**: Conversely, when this heat is dispatched to the outdoors, the receiving environment undergoes an entropy increase of approximately 0.00307 J/K.

These values highlight how energy conservation and entropy are intertwined—energy remains constant, yet entropy dictates spontaneous heat flows, showing a natural tendency toward equilibration.

Energy Conservation

Energy conservation, a fundamental principle of physics, asserts that energy cannot be created or destroyed, only transformed from one form to another. In the context of the Carnot cycle, energy conservation plays a vital role as it dictates the relationship between heat and work in a thermodynamic cycle.

As heat engines like the one described in the exercise operate, they extract heat \( Q_{\text{room}} \) from the cool reservoir, convert part of it into work \( W \), and expel the leftover heat \( Q_{\text{outdoor}} \) to the hot reservoir. Using the law of energy conservation,
\[ Q_{\text{room}} = W + Q_{\text{outdoor}} \]
This equation ensures that all energy forms are accounted for in the process. Our provided solution showed this equation in action by calculating the work done as 0.0553 J and the heat transferred to the outdoors as 0.9447 J. This illustrates how efficiently a cycle can operate while abiding by the inviolable conservation law, also reflecting real-world implications for energy system designs.

Kelvin Temperature Conversion

Temperature conversion is an integral step in any thermodynamic calculation because efficiency and entropy always require absolute temperature scales, like Kelvin.

To convert Celsius to Kelvin, simply add 273.15. This conversion might seem straightforward, but it is paramount for accurate calculations in thermodynamics due to its standardized baseline:

• Temperature of the room: From \(18^{\circ} \text{C} \) to 291.15 K.
• Temperature of the outdoors: From \(35^{\circ} \text{C} \) to 308.15 K.

Using Kelvin ensures consistency, especially because zero Kelvin represents absolute zero, the point at which molecular motion ceases, and this provides a natural and essential reference for heat movement and energy transfer. Accurate conversions facilitate correct efficiencies and changes in entropy computations, as shown in the provided step-by-step solution.