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An object of mass \(m_1\), specific heat \(c_1\), and temperature \(T_1\) is placed in contact with a second object of mass \(m_2\), specific heat \(c_2\), and temperature \(T_2\) > \(T_1\). As a result, the temperature of the first object increases to \(T\) and the temperature of the second object decreases to \(T'\). (a) Show that the entropy increase of the system is $$\Delta S = m_1c_1 ln {T \over T_1} + m_2c_2 ln {T' \over T_2}$$ and show that energy conservation requires that $$m_1c_1 (T - T_1) = m_2c_2 (T_2 - T')$$ (b) Show that the entropy change \(\Delta S\), considered as a function of \(T\), is a \(maximum\) if \(T = T'\), which is just the condition of thermodynamic equilibrium. (c) Discuss the result of part (b) in terms of the idea of entropy as a measure of randomness.

Short Answer

Expert verified
Entropy is maximized at equilibrium when temperatures equalize: \(T = T'\).

Step by step solution

01

Understanding the Problem

We have two objects exchanging heat, and we need to find the change in entropy of the system. Part (a) asks us to derive an expression for the entropy change and confirm energy conservation. Part (b) requires proving that entropy reaches its maximum when temperatures equalize.
02

Apply Energy Conservation

According to the conservation of energy, the energy lost by the hotter object equals the energy gained by the cooler object. Mathematically, this is represented as:\[ m_1c_1 (T - T_1) = m_2c_2 (T_2 - T') \] This equation ensures that all heat exchanged between the objects is accounted for without loss.
03

Derive Entropy Change

The change in entropy \( \Delta S \) for each object can be found using the formula \( \Delta S = mc \ln \frac{T_f}{T_i} \). For the first object, \( \Delta S_1 = m_1c_1 \ln \frac{T}{T_1} \), and for the second object, \( \Delta S_2 = m_2c_2 \ln \frac{T'}{T_2} \). Therefore, the total entropy change is:\[ \Delta S = m_1c_1 \ln \frac{T}{T_1} + m_2c_2 \ln \frac{T'}{T_2} \]
04

Maximize Entropy Change Function

Entropy is a function of the final temperatures \( T \) of the first object and \( T' \) of the second. Differentiate \( \Delta S \) with respect to \( T \) and set \( \frac{d(\Delta S)}{dT} = 0 \) to find critical points. Solving this equation shows that maximum entropy occurs when \( T = T' \), indicating equilibrium.
05

Interpretation of Maximum Entropy

The second law of thermodynamics states that an isolated system will naturally progress towards a state of maximum entropy. In terms of statistical mechanics, higher entropy corresponds to a state in which the system's microstate possibilities are maximized.

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Key Concepts

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

Specific Heat
Specific heat is a property of a substance that describes how much heat is required to change the temperature of a unit mass by one degree Celsius. It plays a crucial role when objects exchange heat. In the exercise, each object has its specific heat, represented by symbols \( c_1 \) and \( c_2 \) for the first and second objects, respectively.
  • Understanding Specific Heat in Context: Imagine you're heating water. Water has a high specific heat, meaning it takes a lot of energy to change its temperature. Similarly, in our exercise, the different specific heats influence how the temperature of each object changes as they exchange heat.
  • Calculating the Heat Exchange: We use the formula \( q = mc\Delta T \) to calculate the heat, where \( q \) is the heat energy, \( m \) is mass, \( c \) is specific heat, and \( \Delta T \) is the change in temperature. This formula is key to understanding how much energy is being transferred as the objects move towards equilibrium.
Specific heat helps us predict the final temperatures and the resultant change in entropy, which are vital for further analysis of the system.
Energy Conservation
Energy conservation is a fundamental principle governing physical processes, stating energy cannot be created or destroyed; it can only change from one form to another. In this context, it means that the total energy exchanged between the two objects remains constant.
  • Energy Balance: The equation \( m_1c_1 (T - T_1) = m_2c_2 (T_2 - T') \) captures this balance. The energy lost by the hotter object (second object) is equal to the energy gained by the cooler object (first object).
  • Application of Conservation: This ensures that all heat transferred is accounted for, confirming the conservation in a closed system. This balance is crucial to accurately calculating entropy changes and understanding the energy dynamics within the system.
  • Implications of Misbalance: If this condition wasn't met, it would imply an external influence or error, violating energy conservation principles, and the calculated entropy change might not reflect the true nature of the interaction.
Using energy conservation helps us understand how the objects' temperatures adjust in pursuit of equilibrium, crucial for maximizing entropy.
Thermodynamic Equilibrium
Thermodynamic equilibrium occurs when two objects in thermal contact no longer exchange heat energy, meaning their temperatures are equal. This condition maximizes the system's entropy.
  • Dynamic Process to Equilibrium: Initially, heat flows from the hotter to the cooler object, causing their temperatures to change until they reach a common temperature,\( T = T' \).
  • Entropy and Randomness: At equilibrium, the system's entropy is at its maximum, indicating a state of highest disorder or randomness. This state signifies that the energy distribution is as spread out as possible.
  • Statistical Interrelation: From a molecular standpoint, equilibrium is when the likelihood of molecules exchanging energy between the two objects is uniform. This reflects how statistical mechanics underpins the concept of equilibrium.
Understanding thermodynamic equilibrium is essential for assessing maximal entropy states. It reflects a balance where no net heat transfer occurs, showcasing a natural tendency towards randomness and disorder.

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Most popular questions from this chapter

(a) Calculate the theoretical efficiency for an Otto-cycle engine with \(\gamma\) = 1.40 and \(r\) = 9.50. (b) If this engine takes in 10,000 J of heat from burning its fuel, how much heat does it discard to the outside air?

A Carnot refrigerator is operated between two heat reservoirs at temperatures of 320 K and 270 K. (a) If in each cycle the refrigerator receives 415 J of heat energy from the reservoir at 270 K, how many joules of heat energy does it deliver to the reservoir at 320 K? (b) If the refrigerator completes 165 cycles each minute, what power input is required to operate it? (c) What is the coefficient of performance of the refrigerator?

An air conditioner operates on 800 W of power and has a performance coefficient of 2.80 with a room temperature of 21.0\(^\circ\)C and an outside temperature of 35.0\(^\circ\)C. (a) Calculate the rate of heat removal for this unit. (b) Calculate the rate at which heat is discharged to the outside air. (c) Calculate the total entropy change in the room if the air conditioner runs for 1 hour. Calculate the total entropy change in the outside air for the same time period. (d) What is the net change in entropy for the system (room + outside air) ?

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