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The already familiar direction of heat transfer from hot to cold is the basis of our first version of the second law of thermodynamics    .

The second law of thermodynamics (first expression)

Heat transfer occurs spontaneously from higher- to lower-temperature bodies but never spontaneously in the reverse direction.

Another way of stating this: It is impossible for any process to have as its sole result heat transfer from a cooler to a hotter object.

Heat engines

Now let us consider a device that uses heat transfer to do work. As noted in the previous section, such a device is called a heat engine, and one is shown schematically in [link] (b). Gasoline and diesel engines, jet engines, and steam turbines are all heat engines that do work by using part of the heat transfer from some source. Heat transfer from the hot object (or hot reservoir) is denoted as Q h size 12{Q rSub { size 8{h} } } {} , while heat transfer into the cold object (or cold reservoir) is Q c size 12{Q rSub { size 8{c} } } {} , and the work done by the engine is W size 12{W} {} . The temperatures of the hot and cold reservoirs are T h size 12{T rSub { size 8{h} } } {} and T c size 12{T rSub { size 8{c} } } {} , respectively.

Part a of the figure shows the spontaneous heat transfer from a hot system to a 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 is shown to flow from hot reservoir to cold reservoir as shown by a bold arrow pointing downward. Part b of the figure shows a heat engine represented as a circle. The hot reservoir at temperature T sub h is represented by a rectangular section at the top and a cold reservoir at temperature T sub c is shown as a rectangular section at the bottom. Heat Q sub h is transferred out of the hot reservoir, work W is the output equals Q sub h minus Q sub c, and heat Q sub c is the heat transferred into the cold reservoir. All these are shown using bold arrows.
(a) Heat transfer occurs spontaneously from a hot object to a cold one, consistent with the second law of thermodynamics. (b) A heat engine, represented here by a circle, uses part of the heat transfer to do work. The hot and cold objects are called the hot and cold reservoirs. Q h size 12{Q rSub { size 8{h} } } {} is the heat transfer out of the hot reservoir, W size 12{W} {} is the work output, and Q c size 12{Q rSub { size 8{c} } } {} is the heat transfer into the cold reservoir.

Because the hot reservoir is heated externally, which is energy intensive, it is important that the work is done as efficiently as possible. In fact, we would like W size 12{W} {} to equal Q h size 12{Q rSub { size 8{h} } } {} , and for there to be no heat transfer to the environment ( Q c = 0 size 12{Q rSub { size 8{c} } =0} {} ). Unfortunately, this is impossible. The second law of thermodynamics    also states, with regard to using heat transfer to do work (the second expression of the second law):

The second law of thermodynamics (second expression)

It is impossible in any system for heat transfer from a reservoir to completely convert to work in a cyclical process in which the system returns to its initial state.

A cyclical process    brings a system, such as the gas in a cylinder, back to its original state at the end of every cycle. Most heat engines, such as reciprocating piston engines and rotating turbines, use cyclical processes. The second law, just stated in its second form, clearly states that such engines cannot have perfect conversion of heat transfer into work done. Before going into the underlying reasons for the limits on converting heat transfer into work, we need to explore the relationships among W size 12{W} {} , Q h size 12{Q rSub { size 8{h} } } {} , and Q c size 12{Q rSub { size 8{c} } } {} , and to define the efficiency of a cyclical heat engine. As noted, a cyclical process brings the system back to its original condition at the end of every cycle. Such a system's internal energy U is the same at the beginning and end of every cycle—that is, Δ U = 0 size 12{ΔU=0} {} . The first law of thermodynamics states that

Δ U = Q W , size 12{ΔU=Q - W} {}

where Q size 12{Q} {} is the net heat transfer during the cycle ( Q = Q h Q c size 12{Q=Q rSub { size 8{h} } - Q rSub { size 8{c} } } {} ) and W size 12{W} {} is the net work done by the system. Since Δ U = 0 size 12{ΔU=0} {} for a complete cycle, we have

Questions & Answers

If a prism is fully imersed in water then the ray of light will normally dispersed or their is any difference?
Anurag Reply
the same behavior thru the prism out or in water bud abbot
Ju
If this will experimented with a hollow(vaccum) prism in water then what will be result ?
Anurag
What was the previous far point of a patient who had laser correction that reduced the power of her eye by 7.00 D, producing a normal distant vision power of 50.0 D for her?
Jaydie Reply
What is the far point of a person whose eyes have a relaxed power of 50.5 D?
Jaydie
What is the far point of a person whose eyes have a relaxed power of 50.5 D?
Jaydie
A young woman with normal distant vision has a 10.0% ability to accommodate (that is, increase) the power of her eyes. What is the closest object she can see clearly?
Jaydie
29/20 ? maybes
Ju
In what ways does physics affect the society both positively or negatively
Princewill Reply
how can I read physics...am finding it difficult to understand...pls help
rerry Reply
try to read several books on phy don't just rely one. some authors explain better than other.
Ju
And don't forget to check out YouTube videos on the subject. Videos offer a different visual way to learn easier.
Ju
hope that helps
Ju
I have a exam on 12 february
David Reply
what is velocity
Jiti
the speed of something in a given direction.
Ju
what is a magnitude in physics
Jiti Reply
Propose a force standard different from the example of a stretched spring discussed in the text. Your standard must be capable of producing the same force repeatedly.
Giovani Reply
What is meant by dielectric charge?
It's Reply
what happens to the size of charge if the dielectric is changed?
Brhanu Reply
omega= omega not +alpha t derivation
Provakar Reply
u have to derivate it respected to time ...and as w is the angular velocity uu will relace it with "thita × time""
Abrar
do to be peaceful with any body
Brhanu Reply
the angle subtended at the center of sphere of radius r in steradian is equal to 4 pi how?
Saeed Reply
if for diatonic gas Cv =5R/2 then gamma is equal to 7/5 how?
Saeed
define variable velocity
Ali Reply
displacement in easy way.
Mubashir Reply
binding energy per nucleon
Poonam Reply
Practice Key Terms 4

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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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