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The entropy of a hot baked potato decreases as it cools. Is this a violation of the increase of entropy principle? Explain.

Short Answer

Expert verified
Answer: No, the decrease in entropy of a hot baked potato as it cools does not violate the entropy increase principle. This is because the overall entropy change in the system, which includes both the potato and its surroundings, still increases. While the potato loses entropy and becomes more ordered, the surroundings gain heat and experience an increase in entropy. As long as the entropy gain of the surroundings is greater than or equal to the entropy loss of the cooling potato, the entropy increase principle is not violated.

Step by step solution

01

Recall the principle of entropy increase

The increase of entropy principle states that in any natural process, the total entropy of an isolated system will either remain the same or increase, never decrease. In order to determine if the cooling potato violates this principle, we need to consider the total entropy change of the potato and its environment as one system.
02

Analyze the entropy changes in the potato and its surroundings

As the hot baked potato cools down, it loses heat to its surroundings. This means the potato's entropy decreases because it is losing thermal energy and becoming more ordered. However, as the surroundings absorb the heat from the potato, they become warmer, leading to an entropy increase. So we should consider the combined entropy change of both the potato and its surroundings.
03

Evaluate the overall entropy change

When the potato loses heat, its entropy decreases, while the surroundings' entropy increases due to gaining heat. According to the Second Law of Thermodynamics, the overall entropy of an isolated system, in this case, the potato and its surroundings combined, can either increase or remain the same. Thus, we need to check if the entropy gain of the surroundings compensates the entropy loss of the potato.
04

Determine the violation of the entropy increase principle

As long as the entropy gain of the surroundings is greater than or equal to the entropy loss of the cooling potato, the overall entropy increase principle is not violated. In most cases, the entropy increase of the surroundings more than compensates for the loss in entropy of the potato, resulting in a net increase in total entropy for the system. Therefore, the decrease in entropy of the cooling potato is not a violation of the entropy increase principle.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Second Law of Thermodynamics
When it comes to understanding the behavior of energy in our universe, the Second Law of Thermodynamics acts as a fundamental guidepost. Simply put, it states that in any given process, the total entropy of an isolated system can increase or remain constant, but it cannot decrease.

Consider a dance between energy and disorder. As systems evolve naturally, they tend to move towards a state of greater chaos or entropy. This might manifest as heat dispersing from a warm object to its cooler surroundings, or gas expanding to fill a vacuum.

This principle is crucial when pondering scenarios like the cooling of a hot baked potato. As it releases heat, the entropy within the potato decreases, suggesting a more ordered state. However, through the lens of the Second Law, the complete picture must be considered – that is, the potato along with its environment. Thus, when we assess the overall disorder of the potato and the air around it, we adhere to the important tenet that entropy, on a whole, is not decreasing, aligning with this unyielding law of thermodynamics.
Isolated Systems
In thermodynamics, an isolated system is akin to an impregnable castle – nothing enters, nothing leaves. It's fully self-contained with no energy or matter crossing its boundaries. In our everyday experience, truly isolated systems are rare; however, they are a useful idealization for understanding the laws of thermodynamics.

Relate this to our baked potato scenario: if we view the potato and its immediate surroundings as an isolated system, it means the heat the potato loses must be gained by the surrounding environment because it has nowhere else to go. Within this isolated system, the key factor to consider is the total entropy change – the sum of what happens inside our 'castle'.

The concept of an isolated system reminds us that sometimes, to really grasp the fate of entropy, we need to step back and look at the bigger picture, beyond just the potato itself.
Entropy Change
Entropy change measures the transition in disorder within a system. To envision this, picture a deck of cards: ordered when new and shuffled into disorder over time. Some processes cause entropy to rise (shuffling the deck), while others result in a decline (sorting the cards back).

In the realm of the cooling baked potato, the potato's entropy decreases as it cools because it is effectively 'sorting its energy cards', becoming more organized. Yet such a localized decrease in entropy, alarming as it may seem, doesn't tell the whole story. It's the entropy change of the entire system, including its environment, that matters. Thermodynamically, when heat is transferred out of the potato and into the surroundings, it's akin to shuffling the environmental 'energy cards' into a state of greater disorder, overall increasing entropy, in line with the Second Law.

Whenever you're faced with the entropy change of an individual component, remember to consider how its environment reacts to the exchange of energy to see if the universal commitment to disorder is upheld.
Heat Transfer
Heat transfer is the journey of thermal energy from one place to another, driven by temperature differences. It's the story of energy on the move – from the radiant warmth of a campfire to the toasty flow from a hot potato to its cooler environs. This movement can occur via conduction, convection, or radiation.

The tale of our cooling potato heavily relies on this concept. As heat transfers from the potato to its surroundings, it speaks to a shift of thermal energy; the environment warms up while the potato itself cools down. This heat transfer is not merely a literal shift in warmth but also a metaphorical transfer of entropy. Whenever heat moves, it carries with it a parcel of disorder, redistributing entropy across the system.

The insightful twist in these heat-driven narratives is that while one object becomes more ordered, the environment inevitably compensates by becoming more disordered, thus preserving the overarching theme of the Second Law of Thermodynamics and ensuring that the principle of entropy increase is never breached.

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Most popular questions from this chapter

Steam enters an adiabatic nozzle at \(2 \mathrm{MPa}\) and \(350^{\circ} \mathrm{C}\) with a velocity of \(55 \mathrm{m} / \mathrm{s}\) and exits at \(0.8 \mathrm{MPa}\) and \(390 \mathrm{m} / \mathrm{s}\). If the nozzle has an inlet area of \(7.5 \mathrm{cm}^{2},\) determine (a) the exit temperature and (b) the rate of entropy generation for this process.

Consider two bodies of identical mass \(m\) and specific heat \(c\) used as thermal reservoirs (source and sink) for a heat engine. The first body is initially at an absolute temperature \(T_{1}\) while the second one is at a lower absolute temperature \(T_{2}\). Heat is transferred from the first body to the heat engine, which rejects the waste heat to the second body. The process continues until the final temperatures of the two bodies \(T_{f}\) become equal. Show that \(T_{f}=\sqrt{T_{1} T_{2}}\) when the heat engine produces the maximum possible work.

Steam is compressed from 6 MPa and \(300^{\circ} \mathrm{C}\) to 10 MPa isentropically. The final temperature of the steam is \((a) 290^{\circ} \mathrm{C}\) \((b) 300^{\circ} \mathrm{C}\) \((c) 311^{\circ} \mathrm{C}\) \((d) 371^{\circ} \mathrm{C}\) \((e) 422^{\circ} \mathrm{C}\)

An ordinary egg can be approximated as a \(5.5-\mathrm{cm}-\) diameter sphere. The egg is initially at a uniform temperature of \(8^{\circ} \mathrm{C}\) and is dropped into boiling water at \(97^{\circ} \mathrm{C}\). Taking the properties of the egg to be \(\rho=1020 \mathrm{kg} / \mathrm{m}^{3}\) and \(c_{p}=3.32 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C},\) determine \((a)\) how much heat is transferred to the egg by the time the average temperature of the egg rises to \(70^{\circ} \mathrm{C}\) and \((b)\) the amount of entropy generation associated with this heat transfer process.

An insulated piston-cylinder device initially contains \(0.02 \mathrm{m}^{3}\) of saturated liquid-vapor mixture of water with a quality of 0.1 at \(100^{\circ} \mathrm{C} .\) Now some ice at \(-18^{\circ} \mathrm{C}\) is dropped into the cylinder. If the cylinder contains saturated liquid at \(100^{\circ} \mathrm{C}\) when thermal equilibrium is established, determine (a) the amount of ice added and ( \(b\) ) the entropy generation during this process. The melting temperature and the heat of fusion of ice at atmospheric pressure are \(0^{\circ} \mathrm{C}\) and \(333.7 \mathrm{kJ} / \mathrm{kg}\).

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