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Chapter 6: Problem 8

How can entropy be transferred into, or out of, a closed system? A control volume?

Short Answer

Expert verified
Entropy can be transferred into or out of a closed system through heat. In a control volume, both heat and mass exchange contribute to entropy transfer.

Step by step solution

01

- Understand entropy transfer in a closed system

In a closed system, entropy can only be transferred through heat interaction. According to the second law of thermodynamics, when heat enters the system, entropy increases. Conversely, when heat leaves the system, entropy decreases. Remember, for closed systems, mass does not cross the system boundary, so only heat transfer affects entropy.
02

- Application to a control volume

In a control volume, entropy transfer can occur through both heat and mass exchange. When mass enters or exits the volume, it carries entropy with it. Additionally, heat transfer across the boundary of the control volume also affects the entropy content. This is because a control volume can exchange both energy and mass with its surroundings.

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Key Concepts

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

Entropy in Closed Systems
Entropy measures disorder or randomness in a system. In a closed system, no mass can cross the boundary, leaving only heat transfer as a means for entropy change.
When heat enters a closed system, the molecules inside become more agitated, increasing the system's entropy. According to the second law of thermodynamics, this happens naturally and spontaneously. Conversely, when heat exits the system, the molecular motion slows down, leading to a decrease in entropy.
Keep in mind:
  • Entropy increases when heat enters the system.
  • Entropy decreases when heat exits the system.
Thus, in a closed system, the control of entropy is tied solely to the heat transfer across its boundaries.
Entropy in Control Volumes
A control volume differs from a closed system in that it can exchange both mass and energy with its surroundings. This means entropy can be transferred through both heat and mass flow.
When mass enters a control volume, it brings its entropy. This is because every bulk of mass contains a certain entropy amount relative to its temperature and pressure. Similarly, when mass leaves the control volume, it takes away entropy.
Key contributions to entropy in control volumes include:
  • Heat transfer: Similar to closed systems, heat entering the control volume increases entropy, while heat leaving decreases it.
  • Mass transfer: Mass entering adds entropy, and mass exiting takes it away.
This dual transfer mechanism makes understanding entropy changes in control volumes more complex compared to closed systems.
Second Law of Thermodynamics
The second law of thermodynamics is a fundamental principle governing the transfer and conversion of energy. One of its crucial statements is that entropy of an isolated system always increases over time.
Here's how it relates to entropy transfer:
  • Natural processes: These tend to move towards increasing disorder or greater entropy. This explains why heat flows spontaneously from hot to cold areas.
  • Closed systems: Heat transfer determines the entropy change, always striving to maintain or increase entropy.
  • Control volumes: Factors are broader here, with both mass and heat transfer influencing entropy.
The second law states that no process is 100% efficient in energy conversion. Some energy will always be lost to increasing entropy, reminding us that energy quality degrades over time in any real-life process.
The law underscores the importance of energy management and efficiency in engineering and everyday applications.

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Most popular questions from this chapter

A patent application describes a device for chilling water. At steady state, the device receives energy by heat transfer at a location on its surface where the temperature is \(540^{\circ} \mathrm{F}\) and discharges energy by heat transfer to the surroundings at another location on its surface where the temperature is \(100^{\circ} \mathrm{F}\). A warm liquid water stream enters at \(100^{\circ} \mathrm{F}, 1 \mathrm{~atm}\) and a cool stream exits at temperature \(T\) and \(1 \mathrm{~atm}\). The device requires no power input to operate, there are no significant effects of kinetic and potential energy, and the water can be modeled as incompressible. Plot the minimum theoretical heat addition required, in Btu per \(\mathrm{lb}\) of cool water exiting the device, versus \(T\) ranging from 60 to \(100^{\circ} \mathrm{F}\).

An electric motor operating at steady state draws a current of 10 amp with a voltage of \(220 \mathrm{~V}\). The output shaft rotates at 1000 RPM with a torque of \(16 \mathrm{~N} \cdot \mathrm{m}\) applied to an external load. The rate of heat transfer from the motor to its surroundings is related to the surface temperature \(T_{\mathrm{b}}\) and the ambient temperature \(T_{0}\) by \(\mathrm{hA}\left(T_{\mathrm{b}}-T_{0}\right)\), where \(\mathrm{h}=100 \mathrm{~W} / \mathrm{m}^{2}\). \(\mathrm{K}, \mathrm{A}=0.195 \mathrm{~m}^{2}\), and \(T_{0}=293 \mathrm{~K}\). Energy transfers are considered positive in the directions indicated by the arrows on Fig. P6.51. (a) Determine the temperature \(T_{\mathrm{b}}\), in \(\mathrm{K}\). (b) For the motor as the system, determine the rate of entropy production, in \(\mathrm{kW} / \mathrm{K}\). (c) If the system boundary is located to take in enough of the nearby surroundings for heat transfer to take place at temperature \(T_{0}\), determine the rate of entropy production, in \(\mathrm{kW} / \mathrm{K}\), for the enlarged system.

Air enters a compressor operating at steady state at \(17^{\circ} \mathrm{C}\), 1 bar and exits at a pressure of 5 bar. Kinetic and potential energy changes can be ignored. If there are no internal irreversibilities, evaluate the work and heat transfer, each in \(\mathrm{kJ}\) per \(\mathrm{kg}\) of air flowing, for the following cases: (a) isothermal compression. (b) polytropic compression with \(n=1.3\). (c) adiabatic compression. Sketch the processes on \(p-v\) and \(T-s\) coordinates and associate areas on the diagrams with the work and heat transfer in each case. Referring to your sketches, compare for these cases the magnitudes of the work, heat transfer, and final temperatures, respectively.

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.

A gas flows through a one-inlet, one-exit control volume operating at steady state. Heat transfer at the rate \(\dot{Q}_{\mathrm{cv}}\) takes place only at a location on the boundary where the temperature is \(T_{\mathrm{b}}\). For each of the following cases, determine whether the specific entropy of the gas at the exit is greater than, equal to, or less than the specific entropy of the gas at the inlet: (a) no internal irreversibilities, \(\dot{Q}_{\mathrm{cv}}=0\). (b) no internal irreversibilities, \(\dot{Q}_{\mathrm{cv}}<0\). (c) no internal irreversibilities, \(\dot{Q}_{\mathrm{cv}}>0\). (d) internal irreversibilities, \(\dot{Q}_{\mathrm{cv}}<0\). (e) internal irreversibilities, \(\dot{Q}_{\mathrm{cv}} \geq 0\)
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