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How To Calculate Change In Entropy Physics

Written By Sunday, April 17, 2022 Add Comment Edit

Learning Objectives

By the end of this section, y'all will be able to:

  • Define entropy.
  • Calculate the increase of entropy in a arrangement with reversible and irreversible processes.
  • Explicate the expected fate of the universe in entropic terms.
  • Calculate the increasing disorder of a organisation.

Photograph shows a glass of a beverage with ice cubes and a straw, placed on a paper napkin on the table. There is a piece of sliced lemon at the edge of the glass. There is condensate around the outside surface of the glass, giving the appearance that the ice is melting.

Figure 1. The ice in this drinkable is slowly melting. Eventually the liquid will reach thermal equilibrium, every bit predicted past the 2d police of thermodynamics. (credit: Jon Sullivan, PDPhoto.org)

There is yet another way of expressing the second constabulary of thermodynamics. This version relates to a concept called entropy. By examining information technology, nosotros shall see that the directions associated with the second constabulary—oestrus transfer from hot to cold, for example—are related to the tendency in nature for systems to get disordered and for less energy to be bachelor for apply every bit piece of work. The entropy of a system can in fact be shown to be a measure of its disorder and of the unavailability of free energy to do work.

Making Connections: Entropy, Energy, and Piece of work

Recall that the elementary definition of energy is the ability to do piece of work. Entropy is a measure out of how much energy is not available to do work. Although all forms of energy are interconvertible, and all can be used to do work, information technology is not always possible, even in principle, to convert the unabridged available energy into piece of work. That unavailable energy is of involvement in thermodynamics, because the field of thermodynamics arose from efforts to convert heat to piece of work.

Nosotros can see how entropy is divers past recalling our discussion of the Carnot engine. Nosotros noted that for a Carnot bike, and hence for any reversible processes,

[latex]\displaystyle\frac{Q_{\text{c}}}{Q_{\text{h}}}=\frac{T_{\text{c}}}{T_{\text{h}}}\\[/latex].

Rearranging terms yields

[latex]\displaystyle\frac{Q_{\text{c}}}{T_{\text{c}}}=\frac{Q_{\text{h}}}{T_{\text{h}}}\\[/latex]

for any reversible process. Q c and Q h are absolute values of the heat transfer at temperatures T c and T h, respectively. This ratio of [latex]\frac{Q}{T}\\[/latex] is divers to be the change in entropy ΔSouth for a reversible process, [latex]\Delta{South}=\left(\frac{Q}{T}\right)_{\text{rev}}\\[/latex], where Q is the oestrus transfer, which is positive for rut transfer into and negative for oestrus transfer out of, and T is the absolute temperature at which the reversible process takes identify. The SI unit for entropy is joules per kelvin (J/K). If temperature changes during the process, then it is usually a good approximation (for small changes in temperature) to take T to be the average temperature, avoiding the need to utilize integral calculus to detect ΔS.

The definition of ΔS is strictly valid only for reversible processes, such as used in a Carnot engine. However, we tin can find ΔS precisely even for real, irreversible processes. The reason is that the entropy S of a system, like internal energy U, depends but on the state of the system and not how it reached that condition. Entropy is a belongings of state. Thus the change in entropy ΔS of a system between state one and state 2 is the aforementioned no affair how the alter occurs. We just need to notice or imagine a reversible process that takes the states from state 1 to land 2 and calculate ΔDue south for that process. That will be the modify in entropy for whatever procedure going from land i to state 2. (Run into Effigy 2.)

The diagram shows a schematic representation of a system that goes from state one with entropy S sub one to state two with entropy S sub two. The two states are shown as two circles drawn a distance apart. Two arrows represent two different processes to take the system from state one to state two. A straight arrow pointing from state one to state two shows a reversible process. The change in entropy for this process is given by delta S equals Q divided by T. The second process is marked as a curving arrow from state one to state two, showing an irreversible process. The change in entropy for this process is given by delta S sub irreversible equals delta S sub reversible equals S sub two minus S sub one.

Figure 2. When a organisation goes from state i to state two, its entropy changes by the same amount ΔSouth, whether a hypothetical reversible path is followed or a real irreversible path is taken.

Now let us accept a look at the alter in entropy of a Carnot engine and its heat reservoirs for 1 full cycle. The hot reservoir has a loss of entropy [latex]\Delta{Due south}_{\text{h}}=\frac{-Q_{\text{h}}}{T_{\text{h}}}\\[/latex], considering heat transfer occurs out of it (recall that when heat transfers out, and so Q has a negative sign). The common cold reservoir has a gain of entropy[latex]\Delta{S}_{\text{c}}=\frac{Q_{\text{c}}}{T_{\text{c}}}\\[/latex], considering heat transfer occurs into it. (We assume the reservoirs are sufficiently big that their temperatures are constant.) So the total alter in entropy is ΔS tot = ΔSouth h + ΔDue south c .

Thus, since we know that [latex]\frac{Q_{\text{h}}}{T_{\text{h}}}=\frac{Q_{\text{c}}}{T_{\text{c}}}\\[/latex] for a Carnot engine, [latex]\Delta{S}_{\text{tot}}=\frac{Q_{\text{h}}}{T_{\text{h}}}=\frac{Q_{\text{c}}}{T_{\text{c}}}=0\\[/latex].

This result, which has general validity, ways that the total change in entropy for a organization in any reversible process is zero.

The entropy of diverse parts of the system may change, but the total change is nix. Furthermore, the system does not touch the entropy of its surroundings, since heat transfer between them does not occur. Thus the reversible process changes neither the total entropy of the organisation nor the entropy of its surroundings. Sometimes this is stated as follows: Reversible processes do not touch the total entropy of the universe. Real processes are not reversible, though, and they practise change full entropy. Nosotros tin can, withal, use hypothetical reversible processes to determine the value of entropy in real, irreversible processes. Example 1 illustrates this point.

Example 1. Entropy Increases in an Irreversible (Real) Procedure

Spontaneous oestrus transfer from hot to cold is an irreversible procedure. Summate the total change in entropy if 4000 J of oestrus transfer occurs from a hot reservoir at T h = 600 K(327ºC) to a cold reservoir at T c = 250 K(−23ºC), bold in that location is no temperature change in either reservoir. (Meet Effigy iii.)

Part a of the figure shows the irreversible heat transfer from the hot system to the cold system. The hot reservoir at temperature T sub h is represented by a rectangular section in the top and the cold reservoir at temperature T sub c is shown as a rectangular section at the bottom. Heat Q is shown to flow from hot reservoir to cold reservoir as shown by a continuous bold arrow pointing downward. The heat is a direct transfer from T sub h to T sub c. The entropy change delta S for an irreversible process is shown equal to entropy change delta S for a reversible process. Part b of the figure shows two reversible heat transfers from the hot system to the cold system. The hot reservoir at temperature T sub h is represented by a rectangular section in the top and the cold reservoir at temperature T sub c is shown as a rectangular section at the bottom. Heat Q is shown to flow out of the hot reservoir, and an equal amount of heat Q is shown to flow into the cold reservoir as shown by two arrows representing two reversible processes and not a direct transfer from T sub h to T sub c. The entropy change delta S for an irreversible process is shown equal to entropy change delta S for a reversible process.

Figure 3. (a) Oestrus transfer from a hot object to a cold one is an irreversible process that produces an overall increase in entropy. (b) The same final state and, thus, the aforementioned change in entropy is achieved for the objects if reversible heat transfer processes occur between the two objects whose temperatures are the same as the temperatures of the respective objects in the irreversible process.

Strategy

How can nosotros calculate the change in entropy for an irreversible process when ΔSouth tot = ΔDue south h + ΔSouthward c is valid simply for reversible processes? Remember that the total alter in entropy of the hot and common cold reservoirs will be the same whether a reversible or irreversible process is involved in oestrus transfer from hot to cold. So we can calculate the modify in entropy of the hot reservoir for a hypothetical reversible process in which 4000 J of heat transfer occurs from it; then we do the aforementioned for a hypothetical reversible procedure in which 4000 J of estrus transfer occurs to the common cold reservoir. This produces the same changes in the hot and common cold reservoirs that would occur if the heat transfer were allowed to occur irreversibly between them, and and then it also produces the same changes in entropy.

Solution

We now calculate the two changes in entropy using ΔDue south tot = ΔS h + ΔS c. First, for the heat transfer from the hot reservoir,

[latex]\displaystyle\Delta{S}_{\text{h}}=\frac{-Q_{\text{h}}}{T_{\text{h}}}=\frac{-4000\text{ J}}{600\text{ K}}=-vi.67\text{ J/K}\\[/latex]

And for the cold reservoir,

[latex]\displaystyle\Delta{South}_{\text{c}}=\frac{-Q_{\text{c}}}{T_{\text{c}}}=\frac{4000\text{ J}}{250\text{ Chiliad}}=sixteen.0\text{ J/K}\\[/latex]

Thus the total is

[latex]\begin{assortment}{lll}\Delta{S}_{\text{tot}}&=&\Delta{S}_{\text{h}}+\Delta{Due south}_{\text{c}}\\\text{ }&=&\left(-6.67+sixteen.0\right)\text{ J/K}\\\text{ }&=&9.33\text{ J/K}\end{assortment}\\[/latex]

Discussion

There is an increase in entropy for the system of two heat reservoirs undergoing this irreversible oestrus transfer. We will see that this means there is a loss of ability to practise work with this transferred energy. Entropy has increased, and free energy has go unavailable to practice work.

It is reasonable that entropy increases for heat transfer from hot to cold. Since the change in entropy is [latex]\frac{Q}{T}\\[/latex], in that location is a larger modify at lower temperatures. The decrease in entropy of the hot object is therefore less than the increase in entropy of the cold object, producing an overall increase, just as in the previous example. This issue is very full general:

There is an increase in entropy for any organisation undergoing an irreversible procedure.

With respect to entropy, there are only two possibilities: entropy is constant for a reversible process, and it increases for an irreversible process. At that place is a fourth version of the second police of thermodynamics stated in terms of entropy:

The full entropy of a organization either increases or remains constant in any procedure; information technology never decreases.

For example, estrus transfer cannot occur spontaneously from common cold to hot, considering entropy would subtract.

Entropy is very different from free energy. Entropy is non conserved but increases in all existent processes. Reversible processes (such equally in Carnot engines) are the processes in which the almost heat transfer to work takes place and are also the ones that go on entropy constant. Thus we are led to make a connectedness between entropy and the availability of free energy to practice work.

Entropy and the Unavailability of Energy to Exercise Work

What does a change in entropy hateful, and why should we be interested in it? Ane reason is that entropy is directly related to the fact that not all estrus transfer can exist converted into piece of work. Example 2 gives some indication of how an increase in entropy results in less heat transfer into work.

Instance 2. Less Piece of work is Produced by a Given Heat Transfer When Entropy Alter is Greater

  1. Calculate the work output of a Carnot engine operating between temperatures of 600 Yard and 100 K for 4000 J of oestrus transfer to the engine.
  2. Now suppose that the 4000 J of estrus transfer occurs commencement from the 600 M reservoir to a 250 K reservoir (without doing whatever work, and this produces the increase in entropy calculated above) earlier transferring into a Carnot engine operating between 250 M and 100 Thou. What work output is produced? (See Figure four.)

Part a of the diagram shows a schematic diagram of a Carnot engine shown as a circle. The hot reservoir is shown as a rectangular section above the circle at temperature T sub h equals six hundred Kelvin. The cold reservoir is shown as a rectangular section below the circle at temperature T sub c equals one hundred Kelvin. A heat Q sub h from the hot reservoir equals four thousand joules is shown to enter the engine as shown as a bold arrow toward the circle from the hot reservoir. A part of it leaves as work W equals three thousand three hundred thirty three joules from the engine. The remaining heat Q sub c equals six hundred sixty seven joules is returned back to the cold reservoir as shown by a bold arrow toward it. Part b of the diagram shows a schematic diagram of a Carnot engine shown as a circle. This engine is shown to have a greater entropy level. An initial heat transfer of four thousand joules occurs from a hot reservoir shown as a rectangular section above the circle toward left at temperature T sub h equals six hundred Kelvin to another rectangular section above the circle at temperature T sub h prime equals two fifty Kelvin. The cold reservoir is shown as a rectangular section below the circle at temperature T sub c prime equals one hundred Kelvin. A heat Q sub h prime from the hot reservoir equals four thousand joules is shown to enter the engine as shown as a bold arrow toward the circle from this hot reservoir. A part of it leaves as work W equals two thousand four hundred joules from the engine. The remaining heat Q sub c equals one thousand six hundred joules is returned back to the cold reservoir as shown by a bold arrow toward it.

Figure 4. (a) A Carnot engine working at between 600 M and 100 1000 has 4000 J of estrus transfer and performs 3333 J of work. (b) The 4000 J of heat transfer occurs first irreversibly to a 250 Thou reservoir and then goes into a Carnot engine. The increase in entropy caused by the heat transfer to a colder reservoir results in a smaller work output of 2400 J. In that location is a permanent loss of 933 J of free energy for the purpose of doing work.

Strategy

In both parts, we must first summate the Carnot efficiency and so the work output.

Solution to Part 1

The Carnot efficiency is given by [latex]\mathit{Eff}_{\text{C}}=1-\frac{T_{\text{c}}}{T_{\text{h}}}\\[/latex].

Substituting the given temperatures yields [latex]\mathit{Eff}_{\text{C}}=1-\frac{100\text{ K}}{600\text{ Yard}}=0.833\\[/latex].

Now the work output tin be calculated using the definition of efficiency for any heat engine as given by [latex]\mathit{Eff}=\frac{W}{Q_{\text{h}}}\\[/latex].

Solving for W and substituting known terms gives

[latex]\brainstorm{array}{lll}West&=&\mathit{Eff}_{\text{C}}Q_{\text{h}}\\\text{ }&=&\left(0.833\right)\left(4000\text{ J}\right)=3333\text{ J}\end{assortment}\\[/latex]

Solution to Part 2

Similarly, [latex]\mathit{Eff}\prime_{\text{C}}=1-\frac{T_{\text{c}}}{T\prime_{\text{c}}}=\frac{100\text{ Thou}}{250\text{ K}}=0.600\\[/latex]

so that

[latex]\begin{array}{lll}Westward&=&\mathit{Eff}\prime_{\text{C}}Q_{\text{h}}\\\text{ }&=&\left(0.600\right)\left(4000\text{ J}\correct)=2400\text{ J}\end{assortment}\\[/latex]

Discussion

At that place is 933 J less work from the aforementioned heat transfer in the second process. This result is of import. The aforementioned estrus transfer into 2 perfect engines produces unlike work outputs, considering the entropy change differs in the two cases. In the second case, entropy is greater and less work is produced. Entropy is associated with the united nationsavailability of free energy to practice piece of work.

When entropy increases, a sure corporeality of energy becomes permanently unavailable to exercise piece of work. The free energy is not lost, merely its character is changed, so that some of it can never be converted to doing piece of work—that is, to an organized strength acting through a distance. For instance, in Instance 2, 933 J less work was done later on an increment in entropy of 9.33 J/Chiliad occurred in the 4000 J heat transfer from the 600 K reservoir to the 250 G reservoir. It tin can be shown that the corporeality of energy that becomes unavailable for work isWestward unavail = ΔSouth⋅T 0, where T 0 is the lowest temperature utilized. In Case 2,W unavail = (nine.33 J/K)(100 Thousand) = 933 J equally institute.

Oestrus Death of the Universe: An Overdose of Entropy

In the early, energetic universe, all matter and free energy were easily interchangeable and identical in nature. Gravity played a vital role in the young universe. Although it may have seemed disorderly, and therefore, superficially entropic, in fact, there was enormous potential energy available to do work—all the future energy in the universe.

As the universe matured, temperature differences arose, which created more opportunity for work. Stars are hotter than planets, for example, which are warmer than icy asteroids, which are warmer still than the vacuum of the space between them.

Most of these are cooling down from their normally violent births, at which time they were provided with energy of their own—nuclear free energy in the case of stars, volcanic energy on Earth and other planets, and then on. Without boosted energy input, nonetheless, their days are numbered.

Every bit entropy increases, less and less free energy in the universe is available to do work. On Earth, we still have great stores of energy such equally fossil and nuclear fuels; big-scale temperature differences, which tin provide wind energy; geothermal energies due to differences in temperature in Earth's layers; and tidal energies owing to our abundance of liquid h2o. As these are used, a sure fraction of the energy they comprise tin can never be converted into doing work. Somewhen, all fuels will be exhausted, all temperatures volition equalize, and it will exist impossible for heat engines to office, or for piece of work to be done.

Entropy increases in a closed organisation, such as the universe. Just in parts of the universe, for instance, in the Solar organization, it is non a locally closed system. Energy flows from the Lord's day to the planets, replenishing Earth'south stores of energy. The Sun volition keep to supply us with energy for most another five billion years. Nosotros volition enjoy directly solar energy, also equally side effects of solar energy, such as air current power and biomass energy from photosynthetic plants. The energy from the Sun volition keep our water at the liquid country, and the Moon'south gravitational pull will continue to provide tidal energy. Just Earth's geothermal energy will slowly run downward and won't exist replenished.

But in terms of the universe, and the very long-term, very large-scale motion picture, the entropy of the universe is increasing, and and so the availability of energy to exercise work is constantly decreasing. Somewhen, when all stars have died, all forms of potential energy have been utilized, and all temperatures have equalized (depending on the mass of the universe, either at a very high temperature following a universal contraction, or a very low one, only before all activity ceases) at that place will exist no possibility of doing work.

Either way, the universe is destined for thermodynamic equilibrium—maximum entropy. This is oft called the heat decease of the universe, and volition mean the finish of all activity. Yet, whether the universe contracts and heats up, or continues to expand and cools down, the cease is non near. Calculations of black holes propose that entropy can easily continue for at least 10100 years.

Lodge to Disorder

Entropy is related not merely to the unavailability of energy to practise work—information technology is too a measure of disorder. This notion was initially postulated by Ludwig Boltzmann in the 1800s. For example, melting a cake of water ice ways taking a highly structured and orderly system of water molecules and converting it into a disorderly liquid in which molecules have no fixed positions. (See Effigy 5.) At that place is a big increase in entropy in the process, as seen in the post-obit example.

The diagram has two images. The first image shows molecules of ice. They are represented as tiny spheres joined to form a floral pattern. The system is shown as ordered. The second image shows what happens when ice melts. The change in entropy delta S is marked between the two images shown by an arrow pointing from first image toward the second image with change in entropy delta S shown greater than zero. The second image represents water shown as tiny spheres moving in a random state. The system is marked as disordered.

Figure 5. When ice melts, it becomes more disordered and less structured. The systematic system of molecules in a crystal construction is replaced by a more random and less orderly movement of molecules without fixed locations or orientations. Its entropy increases because heat transfer occurs into information technology. Entropy is a measure of disorder.

Example 3. Entropy Associated with Disorder

Discover the increase in entropy of 1.00 kg of ice originally at 0º C that is melted to form h2o at 0º C.

Strategy

As before, the change in entropy can be calculated from the definition of ΔSouth one time we discover the energy Q needed to melt the water ice.

Solution

The change in entropy is defined equally: [latex]\Delta{S}=\frac{Q}{T}\\[/latex].

Hither Q is the heat transfer necessary to melt 1.00 kg of ice and is given byQ =mL f, where thousand is the mass and L f is the latent oestrus of fusion. 50 f = 334 kJ/kg for water, so thatQ = (i.00 kg)(334 kJ/kg) = 3.34 × 10v J.

Now the change in entropy is positive, since heat transfer occurs into the ice to cause the stage alter; thus,

[latex]\displaystyle\Delta{Due south}=\frac{Q}{T}=\frac{3.34\times10^5\text{ J}}{T}\\[/latex]

T is the melting temperature of ice. That is, T= 0ºC = 273 M. So the change in entropy is

[latex]\begin{array}{lll}\Delta{South}&=&\frac{3.34\times10^5\text{ J}}{273\text{ K}}\\\text{ }&=&1.22\times10^3\text{ J/K}\end{array}\\[/latex]

Discussion

This is a significant increment in entropy accompanying an increase in disorder.

In some other easily imagined instance, suppose we mix equal masses of h2o originally at ii different temperatures, say 20.0ºC and 40.0ºC. The result is water at an intermediate temperature of xxx.0ºC. Three outcomes have resulted: entropy has increased, some energy has become unavailable to do work, and the system has go less orderly. Let united states of america call back virtually each of these results.

Beginning, entropy has increased for the aforementioned reason that information technology did in Instance 3. Mixing the two bodies of water has the same result as heat transfer from the hot ane and the same estrus transfer into the cold one. The mixing decreases the entropy of the hot water but increases the entropy of the cold water by a greater amount, producing an overall increase in entropy.

Second, once the two masses of water are mixed, at that place is only ane temperature—you cannot run a heat engine with them. The energy that could accept been used to run a heat engine is now unavailable to practise work.

Third, the mixture is less orderly, or to use some other term, less structured. Rather than having two masses at unlike temperatures and with different distributions of molecular speeds, we now take a single mass with a uniform temperature.

These three results—entropy, unavailability of energy, and disorder—are non only related but are in fact substantially equivalent.

Life, Development, and the Second Law of Thermodynamics

Some people misunderstand the second law of thermodynamics, stated in terms of entropy, to say that the process of the evolution of life violates this law. Over time, complex organisms evolved from much simpler ancestors, representing a large decrease in entropy of the Earth's biosphere. It is a fact that living organisms have evolved to exist highly structured, and much lower in entropy than the substances from which they grow. But it is always possible for the entropy of ane part of the universe to decrease, provided the total change in entropy of the universe increases. In equation class, we can write this as ΔS tot = ΔS syst + ΔSouthward envir > 0.

Thus ΔSouthward syst can exist negative as long as ΔS envir is positive and greater in magnitude.

How is it possible for a system to decrease its entropy? Energy transfer is necessary. If I pick upwards marbles that are scattered about the room and put them into a loving cup, my piece of work has decreased the entropy of that system. If I gather iron ore from the ground and catechumen information technology into steel and build a bridge, my work has decreased the entropy of that arrangement. Energy coming from the Dominicus tin can subtract the entropy of local systems on Globe—that is, ΔDue south syst is negative. But the overall entropy of the residue of the universe increases by a greater amount—that is, ΔSouth envir is positive and greater in magnitude. Thus, ΔS tot = ΔS syst + ΔS envir > 0, and the second constabulary of thermodynamics is not violated.

Every time a found stores some solar energy in the form of chemic potential energy, or an updraft of warm air lifts a soaring bird, the Earth tin can exist viewed as a heat engine operating between a hot reservoir supplied past the Sun and a common cold reservoir supplied past night outer space—a heat engine of loftier complexity, causing local decreases in entropy as information technology uses part of the heat transfer from the Sunday into deep space. There is a large total increase in entropy resulting from this massive estrus transfer. A pocket-sized part of this heat transfer is stored in structured systems on Earth, producing much smaller local decreases in entropy. (Come across Figure 6.)

The figure shows the schematic diagram for heat transfer from the Sun into deep space. The picture of the Sun is shown at the left most end of the diagram. The temperature of the Sun is marked as T sub h. The heat Q is shown to flow as a bold arrow pointing till the right end of the diagram which is labeled as deep space. The temperature here is shown as T sub c equals three Kelvin. The Earth is shown as a sphere at the middle of this bold arrow stream between Sun and deep space. The Earth is shown to receive an internal energy delta U. The change in entropy of Earth delta S is shown to be less than zero with a question mark.

Figure 6. Globe's entropy may decrease in the procedure of intercepting a modest part of the heat transfer from the Dominicus into deep space. Entropy for the entire procedure increases greatly while Globe becomes more structured with living systems and stored energy in diverse forms.

PhET Explorations: Reversible Reactions

Watch a reaction proceed over time. How does full energy affect a reaction rate? Vary temperature, barrier summit, and potential energies. Record concentrations and fourth dimension in order to excerpt rate coefficients. Do temperature dependent studies to extract Arrhenius parameters. This simulation is best used with teacher guidance because it presents an analogy of chemical reactions.

Reversible Reactions screenshot.

Click to download the simulation. Run using Coffee.

Section Summary

  • Entropy is the loss of free energy available to do work.
  • Another form of the 2nd constabulary of thermodynamics states that the total entropy of a system either increases or remains constant; it never decreases.
  • Entropy is zero in a reversible process; it increases in an irreversible process.
  • The ultimate fate of the universe is likely to be thermodynamic equilibrium, where the universal temperature is constant and no free energy is available to do work.
  • Entropy is also associated with the tendency toward disorder in a airtight organisation.

Conceptual Questions

  1. A adult female shuts her summer cottage up in September and returns in June. No one has entered the cottage in the concurrently. Explain what she is likely to find, in terms of the 2nd law of thermodynamics.
  2. Consider a system with a sure energy content, from which nosotros wish to extract as much work as possible. Should the system'southward entropy be loftier or depression? Is this orderly or disorderly? Structured or uniform? Explain briefly.
  3. Does a gas go more orderly when it liquefies? Does its entropy change? If and so, does the entropy increment or decrease? Explain your answer.
  4. Explain how water'southward entropy can decrease when information technology freezes without violating the second law of thermodynamics. Specifically, explain what happens to the entropy of its surroundings.
  5. Is a uniform-temperature gas more or less orderly than ane with several different temperatures? Which is more structured? In which can oestrus transfer result in work done without oestrus transfer from another organization?
  6. Give an case of a spontaneous process in which a system becomes less ordered and energy becomes less available to exercise work. What happens to the system's entropy in this procedure?
  7. What is the change in entropy in an adiabatic process? Does this imply that adiabatic processes are reversible? Tin a procedure be precisely adiabatic for a macroscopic system?
  8. Does the entropy of a star increment or subtract every bit it radiates? Does the entropy of the space into which it radiates (which has a temperature of about 3 1000) increment or decrease? What does this practice to the entropy of the universe?
  9. Explain why a building made of bricks has smaller entropy than the same bricks in a disorganized pile. Do this by considering the number of ways that each could be formed (the number of microstates in each macrostate).

Problems & Exercises

  1. (a) On a wintertime 24-hour interval, a certain firm loses 5.00 × 108 J of heat to the outside (about 500,000 Btu). What is the full change in entropy due to this heat transfer alone, assuming an average indoor temperature of 21.0ºC and an average outdoor temperature of five.00ºC? (b) This big change in entropy implies a large amount of energy has become unavailable to do work. Where do we find more free energy when such energy is lost to us?
  2. On a hot summertime solar day, 4.00 × 106 J of heat transfer into a parked car takes place, increasing its temperature from 35.0ºC to 45.0ºC. What is the increase in entropy of the motorcar due to this oestrus transfer alone?
  3. A hot rock ejected from a volcano's lava fountain cools from 1100ºC to 40.0ºC, and its entropy decreases past 950 J/K. How much estrus transfer occurs from the rock?
  4. When 1.60 × 105 J of heat transfer occurs into a meat pie initially at 20.0ºC, its entropy increases by 480 J/K. What is its final temperature?
  5. The Sun radiates energy at the charge per unit of 3.fourscore × x26 West from its 5500ºC surface into dark empty infinite (a negligible fraction radiates onto Earth and the other planets). The constructive temperature of deep space is −270ºC. (a) What is the increment in entropy in one twenty-four hour period due to this heat transfer? (b) How much piece of work is made unavailable?
  6. (a) In reaching equilibrium, how much heat transfer occurs from ane.00 kg of water at forty.0ºC when it is placed in contact with i.00 kg of xx.0ºC water in reaching equilibrium? (b) What is the change in entropy due to this heat transfer? (c) How much work is made unavailable, taking the lowest temperature to exist 20.0ºC? Explicitly show how y'all follow the steps in the Trouble-Solving Strategies for Entropy.
  7. What is the subtract in entropy of 25.0 thou of water that condenses on a bathroom mirror at a temperature of 35.0ºC, assuming no change in temperature and given the latent heat of vaporization to exist 2450 kJ/kg?
  8. Observe the increase in entropy of 1.00 kg of liquid nitrogen that starts at its boiling temperature, boils, and warms to 20.0ºC at constant pressure level.
  9. A large electric power station generates 1000 MW of electricity with an efficiency of 35.0%. (a) Summate the rut transfer to the ability station, Q h, in one twenty-four hour period. (b) How much heat transfer Q c occurs to the environment in one day? (c) If the heat transfer in the cooling towers is from 35.0ºC water into the local air mass, which increases in temperature from eighteen.0ºC to xx.0ºC, what is the full increase in entropy due to this rut transfer? (d) How much energy becomes unavailable to practise work considering of this increment in entropy, assuming an eighteen.0ºC lowest temperature? (Role of Q c could be utilized to operate heat engines or for simply heating the surround, merely it rarely is.)
  10. (a) How much oestrus transfer occurs from 20.0 kg of xc.0ºC h2o placed in contact with 20.0 kg of 10.0ºC water, producing a final temperature of l.0ºC? (b) How much piece of work could a Carnot engine do with this heat transfer, assuming it operates between two reservoirs at constant temperatures of 90.0ºC and x.0ºC? (c) What increase in entropy is produced by mixing xx.0 kg of xc.0ºC h2o with 20.0 kg of ten.0ºC water? (d) Calculate the amount of work fabricated unavailable by this mixing using a depression temperature of ten.0ºC, and compare it with the work done past the Carnot engine. Explicitly show how y'all follow the steps in the Problem-Solving Strategies for Entropy. (e) Discuss how everyday processes brand increasingly more energy unavailable to practice piece of work, equally implied by this problem.

Glossary

entropy: a measurement of a system's disorder and its disability to do piece of work in a organization

modify in entropy: the ratio of estrus transfer to temperature [latex]\frac{Q}{T}\\[/latex]

second law of thermodynamics stated in terms of entropy: the total entropy of a organization either increases or remains constant; it never decreases

Selected Solutions to Problems & Exercises

1. (a) 9.78 × teniv J/K; (b) In order to proceeds more energy, nosotros must generate it from things inside the firm, like a heat pump, human being bodies, and other appliances. As yous know, we use a lot of free energy to keep our houses warm in the wintertime considering of the loss of rut to the outside.

three. viii.01 × 10v J

5. (a) 1.04 × 1031 J/Thou; (b) three.28 × x31 J

7. 199 J/K

9. (a) two.47 × 1014 J; (b) 1.60 × x14 J; (c) 2.85 × tenten J/Thou; (d) 8.29 × x12 J

How To Calculate Change In Entropy Physics,

Source: https://courses.lumenlearning.com/physics/chapter/15-6-entropy-and-the-second-law-of-thermodynamics-disorder-and-the-unavailability-of-energy/

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