/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 13 An isolated system of total mass... [FREE SOLUTION] | 91Ó°ÊÓ

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An isolated system of total mass \(m\) is formed by mixing two equal masses of the same liquid initially at the temperatures \(T_{1}\) and \(T_{2}\). Eventually, the system attains an equilibrium state. Each mass is incompressible with constant specific heat \(c\). (a) Show that the amount of entropy produced is $$ \sigma=m c \ln \left[\frac{T_{1}+T_{2}}{2\left(T_{1} T_{2}\right)^{1 / 2}}\right] $$ (b) Demonstrate that \(\sigma\) must be positive.

Short Answer

Expert verified
\(\sigma = m c \, \ln\left(\frac{T_1 + T_2}{2 (T_1 T_2)^{1/2}}\right)\) is derived and shown to be positive.

Step by step solution

01

Determine Final Equilibrium Temperature

Since the system eventually reaches equilibrium, the final equilibrium temperature can be calculated as the average of the initial temperatures. Therefore, the final temperature, denoted as \(T_f\), is \[T_f = \frac{T_1 + T_2}{2}.\]
02

Calculate Change in Entropy for Each Mass

For the first mass initially at \(T_1\), the change in entropy \(\Delta S_1\) is calculated as \[ \Delta S_1 = m c \, \ln\left(\frac{T_f}{T_1}\right), \] where \(m\) is the mass and \(c\) is the specific heat. Similarly, for the second mass initially at \(T_2\), the change in entropy \(\Delta S_2\) is \[ \Delta S_2 = m c \, \ln\left(\frac{T_f}{T_2}\right). \]
03

Total Entropy Change

The total entropy change \(\Delta S_{total}\) of the system is the sum of the individual entropy changes: \[ \Delta S_{total} = \Delta S_1 + \Delta S_2 = m c \, \ln\left(\frac{T_f}{T_1}\right) + m c \, \ln\left(\frac{T_f}{T_2}\right). \]
04

Simplify the Total Entropy Change

Combine the logarithms using the properties of logarithms: \[ \Delta S_{total} = m c \, \ln\left(\frac{T_f}{T_1} \cdot \frac{T_f}{T_2}\right) = m c \, \ln\left(\frac{T_f^2}{T_1 T_2}\right). \]
05

Substitute the Final Temperature

Substitute \(T_f = \frac{T_1 + T_2}{2}\) into the equation: \[ \Delta S_{total} = m c \, \ln\left(\frac{\left(\frac{T_1 + T_2}{2}\right)^2}{T_1 T_2}\right) = m c \, \ln\left(\frac{(T_1 + T_2)^2}{4 T_1 T_2}\right). \]
06

Final Expression for Entropy Produced

Simplify further to get the final expression for the entropy produced: \[ \sigma = m c \, \ln\left(\frac{T_1 + T_2}{2 (T_1 T_2)^{1/2}}\right). \]
07

Demonstrate that Entropy Change is Positive

Since the argument of the logarithm in \(\sigma\) is always greater than 1 (by the arithmetic mean–geometric mean inequality), \(\ln\left(\frac{T_1 + T_2}{2 (T_1 T_2)^{1/2}}\right) > 0\). Therefore, \(\sigma\) is positive.

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Key Concepts

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

isolated systems
An isolated system in thermodynamics is one that does not exchange matter or energy with its surroundings. This means no heat, work, or mass transfer occurs across its boundaries. In our exercise, the isolated system comprises two equal masses of liquid with different initial temperatures. These masses are mixed, and over time, they reach a common, equilibrium temperature. Understanding isolated systems is crucial because it simplifies our calculations by ensuring that the total energy remains constant. This makes it easier to predict the final temperature and entropy changes.
specific heat
Specific heat, often denoted as 'c', is a property of a substance that indicates how much energy is required to raise the temperature of a unit mass by one unit of temperature. For example, the specific heat of water is about 4.18 J/(g·°C), meaning it takes 4.18 joules to raise the temperature of one gram of water by one degree Celsius. In our exercise, each mass of the liquid has a constant specific heat 'c'. This uniformity simplifies our entropy calculations, as we can apply the same value of 'c' for both masses when calculating their individual entropy changes.
entropy production
Entropy production is an essential concept in thermodynamics, related to the second law, which states that the entropy of an isolated system always increases over time. In our exercise, we calculate the entropy change (or entropy production) that occurs as the two masses of liquid reach thermal equilibrium. The formula derived, \(\sigma = m c \, \ln\left(\frac{T_1 + T_2}{2 \left(T_1 T_2\right)^{1/2}}\right)\), shows us how entropy is generated as the temperatures equalize. Importantly, it confirms that the system's entropy production is positive, which aligns with the second law of thermodynamics. This positive entropy production means that energy has become more spread out or disordered as the system reaches equilibrium.
thermodynamic equilibrium
Thermodynamic equilibrium is a state in which all macroscopic flows of matter and energy have ceased, and the properties of the system are uniform throughout. In our exercise, the two masses at different initial temperatures eventually reach the same final temperature, \(T_f = \frac{T_1 + T_2}{2}\). Achieving thermodynamic equilibrium means that no further changes in temperature or state will occur without external interference. This state is characterized by maximum entropy, as any differences in temperature (a form of energy gradient) have been equalized, leading to an even distribution of energy within the system.

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Most popular questions from this chapter

The temperature of a 12 -oz \((0.354-\mathrm{L})\) can of soft drink is reduced from 20 to \(5^{\circ} \mathrm{C}\) by a refrigeration cycle. The cycle receives energy by heat transfer from the soft drink and discharges energy by heat transfer at \(20^{\circ} \mathrm{C}\) to the surroundings. There are no other heat transfers. Determine the minimum theoretical work input required by the cycle, in \(\mathrm{kJ}\), assuming the soft drink is an incompressible liquid with the properties of liquid water. Ignore the aluminum can.

Water vapor enters an insulated nozzle operating at steady state at \(0.7 \mathrm{MPa}, 320^{\circ} \mathrm{C}, 35 \mathrm{~m} / \mathrm{s}\) and expands to \(0.15 \mathrm{MPa}\). If the isentropic nozzle efficiency is \(94 \%\), determine the velocity at the exit, in \(\mathrm{m} / \mathrm{s}\).

Air is compressed in an axial-flow compressor operating at steady state from \(27^{\circ} \mathrm{C}, 1\) bar to a pressure of \(2.1\) bar. The work input required is \(94.6 \mathrm{~kJ}\) per \(\mathrm{kg}\) of air flowing through the compressor. Heat transfer from the compressor occurs at the rate of \(14 \mathrm{~kJ}\) per \(\mathrm{kg}\) at a location on the compressor's surface where the temperature is \(40^{\circ} \mathrm{C}\). Kinetic and potential energy changes can be ignored. Determine (a) the temperature of the air at the exit, in \({ }^{\circ} \mathrm{C}\). (b) the rate at which entropy is produced within the compressor, in \(\mathrm{kJ} / \mathrm{K}\) per \(\mathrm{kg}\) of air flowing.

A compressor operating at steady state takes in atmospheric air at \(20^{\circ} \mathrm{C}, 1\) bar at a rate of \(1 \mathrm{~kg} / \mathrm{s}\) and discharges air at 5 bar. Plot the power required, in \(\mathrm{kW}\), and the exit temperature, in \({ }^{\circ} \mathrm{C}\), versus the isentropic compressor efficiency ranging from 70 to \(100 \%\). Assume the ideal gas model for the air and neglect heat transfer with the surroundings and changes in kinetic and potential energy.

Air enters a compressor operating at steady state with a volumetric flow rate of \(8 \mathrm{~m}^{3} / \mathrm{min}\) at \(23^{\circ} \mathrm{C}, 0.12 \mathrm{MPa}\). The air is compressed isothermally without internal irreversibilities, exiting at \(1.5 \mathrm{MPa}\). Kinetic and potential energy effects can be ignored. Evaluate the work required and the heat transfer, each in \(\mathrm{kW}\).

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