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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.
(a) Heat 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 same 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 corresponding objects in the irreversible process.

It is reasonable that entropy increases for heat transfer from hot to cold. Since the change in entropy is Q / T size 12{Q/T} {} , there is a larger change 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 result is very general:

There is an increase in entropy for any system undergoing an irreversible process.

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

The total entropy of a system either increases or remains constant in any process; it never decreases.

For example, heat transfer cannot occur spontaneously from cold to hot, because entropy would decrease.

Entropy is very different from energy. Entropy is not conserved but increases in all real processes. Reversible processes (such as in Carnot engines) are the processes in which the most heat transfer to work takes place and are also the ones that keep entropy constant. Thus we are led to make a connection between entropy and the availability of energy to do work.

Entropy and the unavailability of energy to do work

What does a change in entropy mean, and why should we be interested in it? One reason is that entropy is directly related to the fact that not all heat transfer can be converted into work. The next example gives some indication of how an increase in entropy results in less heat transfer into work.

Less work is produced by a given heat transfer when entropy change is greater

(a) Calculate the work output of a Carnot engine operating between temperatures of 600 K and 100 K for 4000 J of heat transfer to the engine. (b) Now suppose that the 4000 J of heat transfer occurs first from the 600 K reservoir to a 250 K reservoir (without doing any work, and this produces the increase in entropy calculated above) before transferring into a Carnot engine operating between 250 K and 100 K. What work output is produced? (See [link] .)

Strategy

In both parts, we must first calculate the Carnot efficiency and then the work output.

Solution (a)

The Carnot efficiency is given by

Eff C = 1 T c T h . size 12{ ital "Eff" rSub { size 8{c} } =1- { {T rSub { size 8{c} } } over {T rSub { size 8{h} } } } } {}

Substituting the given temperatures yields

Eff C = 1 100 K 600 K = 0 . 833 . size 12{ ital "Eff" rSub { size 8{C} } =1- { {"100"" K"} over {"600 K"} } =0 "." "833"} {}

Now the work output can be calculated using the definition of efficiency for any heat engine as given by

Eff = W Q h . size 12{ ital "Eff"= { {W} over {Q rSub { size 8{h} } } } } {}

Solving for W size 12{W} {} and substituting known terms gives

W = Eff C Q h = ( 0.833 ) ( 4000 J ) = 3333 J. alignl { stack { size 12{W= ital "Eff" rSub { size 8{C} } cdot Q rSub { size 8{h} } } {} #" "= \( 0 "." "833" \) \( "4000"" J" \) "=3333 J" "." {} } } {}

Solution (b)

Similarly,

Eff C = 1 T c T′ c = 1 100 K 250 K = 0.600,

so that

W = Eff C Q h = ( 0.600 ) ( 4000 J ) = 2400 J.

Discussion

There is 933 J less work from the same heat transfer in the second process. This result is important. The same heat transfer into two perfect engines produces different work outputs, because 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 un availability of energy to do work.

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.
(a) A Carnot engine working at between 600 K and 100 K has 4000 J of heat transfer and performs 3333 J of work. (b) The 4000 J of heat transfer occurs first irreversibly to a 250 K 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. There is a permanent loss of 933 J of energy for the purpose of doing work.
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Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
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