Find an expression for the change in entropy when two blocks of the same substance and of equal mass, one at the temperature \(T_{\mathrm{h}}\) and the other at \(T_{c}\), are brought into thermal contact and allowed to reach equilibrium. Evaluate the change for two blocks of copper, each of mass \(500 \mathrm{~g}\), with \(C_{p, m}=24.4 \mathrm{~J} \mathrm{~K}^{-1}\) \(\mathrm{mol}^{-1}\), taking \(T_{\mathrm{h}}=500 \mathrm{~K}\) and \(T_{\mathrm{c}}=250 \mathrm{~K}\)

Short Answer

Expert verified
The final equilibrium temperature (T_f) is 375 K. The change in entropy for the hot block is -2.11 kJ/K, for the cold block is 2.11 kJ/K, and the total change in entropy for the system is 0 kJ/K.

Step by step solution

01

Determine the final equilibrium temperature

Assuming no loss of heat to the surroundings and since the masses of the blocks are the same, the final equilibrium temperature, T_f, is the average of T_h and T_c, using the formula: T_f = (T_h + T_c) / 2.
02

Calculate the change in entropy for the hot block

For the hot block cooling down, the change in entropy (deltaS_h) can be calculated using the formula deltaS_h = m * Cp * ln(T_f / T_h), where m is the mass of the block, Cp is the specific heat capacity at constant pressure, and T_f and T_h are the final and initial temperatures of the hot block.
03

Calculate the change in entropy for the cold block

For the cold block heating up, the change in entropy (deltaS_c) can be calculated using the formula deltaS_c = m * Cp * ln(T_f / T_c), where T_c is the initial temperature of the cold block.
04

Calculate the total change in entropy

The total change in entropy (DeltaS_total) for the system is the sum of the changes for each block, i.e., DeltaS_total = deltaS_h + deltaS_c.
05

Evaluate the change in entropy for copper blocks

Substitute the given values: m = 500g, Cp = 24.4 J/K*mol, T_h = 500 K, T_c = 250 K, into the entropy formulas to find deltaS_h and deltaS_c, and then sum them to find DeltaS_total.

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Key Concepts

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

Thermodynamics
Thermodynamics is a branch of physics that deals with the relations between heat and other forms of energy. It describes how thermal energy is converted to and from other forms of energy and how it affects matter. The key to understanding thermodynamics lies in its four laws which explain concepts such as energy conservation, the direction of heat transfer, and the absolute limits of efficiency in energy exchanges.

In the context of our exercise, we deal with the first law of thermodynamics, which is essentially the law of energy conservation. This law states that energy cannot be created or destroyed, only converted from one form to another. When two blocks of different temperatures are allowed to come into contact, energy will transfer from the hotter block to the colder one until they reach a state of thermal equilibrium.

The second law of thermodynamics is particularly relevant for calculating entropy changes. It states that the total entropy of an isolated system can never decrease over time. In our example, the heat transfer results in an increase in entropy, demonstrating how disorder increases in a spontaneous process - moving from a non-equilibrium to an equilibrium state.
Specific Heat Capacity
Specific heat capacity, often shortened to specific heat, is a material-dependent property that indicates the amount of heat required to change the temperature of a unit mass of substance by one degree Celsius (or Kelvin). It is denoted by the symbol 'Cp' when measured at constant pressure.

Understanding specific heat is crucial for solving thermodynamics problems involving temperature changes. The specific heat of a substance tells us how 'reluctant' it is to change its temperature. For example, water has a high specific heat, meaning it requires a lot of energy to heat up or cool down. In our exercise, the specific heat capacity of copper is required to determine how much the entropy changes when the block's temperature changes.

To calculate the change in entropy of the copper blocks, we need to know the specific heat capacity of copper, which allows us to link the energy exchange to temperature change and subsequently to entropy change.
Equilibrium Temperature
The equilibrium temperature is the temperature at which two substances in contact with each other and isolated from external influences will no longer exchange heat, meaning they have reached a state of thermal equilibrium. At this point, there is no net flow of thermal energy between the substances and their temperatures are equal.

Calculating the equilibrium temperature is an essential step when determining the entropy change of a system. As per the exercise, we calculate the final equilibrium temperature by taking the average of the initial temperatures of both copper blocks. This simple approach works because the blocks are of equal mass and made of the same material, hence their heat capacities are equal.

This temperature represents a state of maximum entropy for the system of two blocks, as it is the most disordered state achievable by just the transfer of heat between the two blocks without doing work on the environment or receiving work.
Entropy Formula
Entropy is a measure of the disorder or randomness in a system and is a central concept in the second law of thermodynamics. It quantifies the number of ways a system can be arranged, often interpreted as the degree of uncertainty or unpredictability.

The formula to calculate the change in entropy, \( \Delta S \), due to a temperature change is given by \( \Delta S = m \cdot Cp \cdot \ln(T_f / T_i) \), where \( m \) is the mass of the substance, \( Cp \) is its specific heat capacity, \( T_f \) is the final equilibrium temperature, and \( T_i \) is the initial temperature of the substance.

In our exercise, we apply this formula separately to each copper block, one initially at a higher temperature and one initially at a lower temperature. Aftermath, we add the entropy change of each block to find the total entropy change for the system. The logarithmic nature of the entropy formula reflects the idea that entropy is related to the exponential increase in the number of microstates accessible to a system as its temperature increases.

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Most popular questions from this chapter

Suppose that \(S\) is regarded as a function of \(p\) and \(\mathrm{T}\). Show that \(T \mathrm{dS}=\) \(C_{\mathrm{p}} \mathrm{d} T-\alpha T V \mathrm{~d} p .\) Hence, show that the energy transferred as heat when the pressure on an incompressible liquid or solid is increased by \(\Delta p\) is equal to \(-\alpha T V \Delta p .\) Evaluate \(q\) when the pressure acting on \(100 \mathrm{~cm}^{3}\) of mercury at \(0^{\circ} \mathrm{C}\) is increased by \(1.0\) kbar. \(\left(\alpha=1.82 \times 10^{-4} \mathrm{~K}^{-1} .\right)\)

Prove that two reversible adiabatic paths can never cross. Assume that the energy of the system under consideration is a function of temperature only. (Hint. Suppose that two such paths can intersect, and complete a cycle with the two paths plus one isothermal path. Consider the changes accompanying each stage of the cycle and show that they conflict with the Kelvin statement of the Second Law.)

The cycle involved in the operation of an internal combustion engine is called the Otto cycle. Air can be considered to be the working substance and can be assumed to be a perfect gas. The cycle consists of the following steps: (1) reversible adiabatic compression from \(\mathrm{A}\) to \(\mathrm{B},(2)\) reversible constantvolume pressure increase from \(\mathrm{B}\) to \(\mathrm{C}\) due to the combustion of a small amount of fuel, (3) reversible adiabatic expansion from \(\mathrm{C}\) to \(\mathrm{D}\), and \((4)\) reversible and constant-volume pressure decrease back to state A. Determine the change in entropy (of the system and of the surroundings) tor each step of the cycle and determine an expression for the efficiency of the cycle, assuming that the heat is supplied in Step \(2 .\) Evaluate the efficiency for a compression ratio of \(10: 1\). Assume that, in state \(\mathrm{A}, \mathrm{V}=4.00 \mathrm{dm}^{3}, p=1.00 \mathrm{~atm}\), and \(\mathrm{T}=300 \mathrm{~K}\), that \(\mathrm{V}_{\mathrm{A}}==10 \mathrm{~V}_{\mathrm{B}}, \mathrm{p}_{\mathrm{C}} / \mathrm{p}_{\mathrm{B}}=5\), and that \(C_{p m}=\frac{2}{2} R\).

A gaseous sample consisting of \(1.00 \mathrm{~mol}\) molecules is described by the equation of state \(p \mathrm{~V}_{\mathrm{m}}=\mathrm{RT}(1+B p)\). Initially at \(373 \mathrm{~K}\), it undergoes JouleThomson expansion from \(100 \mathrm{~atm}\) to \(1.00 \mathrm{~atm}\). Given that \(C_{p, \mathrm{~m}}=R, \mu=0.21\) \(\mathrm{K} \mathrm{atm}^{-1}, B=-0.525(\mathrm{~K} / T) \mathrm{atm}^{-1}\), and that these are constant over the temperature range involved, calculate \(\Delta T\) and \(\Delta S\) for the gas.

Show that, for a perfect gas, \((\partial U / \partial S)_{v}=T\) and \((\partial U / \partial V)_{\mathrm{S}}=-p\).

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