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By the end of this section, you will be able to:
  • Contrast the second law of thermodynamics statements according to Kelvin and Clausius formulations
  • Interpret the second of thermodynamics via irreversibility

Earlier in this chapter, we introduced the Clausius statement of the second law of thermodynamics, which is based on the irreversibility of spontaneous heat flow. As we remarked then, the second law of thermodynamics can be stated in several different ways, and all of them can be shown to imply the others. In terms of heat engines, the second law of thermodynamics may be stated as follows:

Second law of thermodynamics (kelvin statement)

It is impossible to convert the heat from a single source into work without any other effect.

This is known as the Kelvin statement of the second law of thermodynamics    . This statement describes an unattainable “ perfect engine    ,” as represented schematically in [link] (a). Note that “without any other effect” is a very strong restriction. For example, an engine can absorb heat and turn it all into work, but not if it completes a cycle . Without completing a cycle, the substance in the engine is not in its original state and therefore an “other effect” has occurred. Another example is a chamber of gas that can absorb heat from a heat reservoir and do work isothermally against a piston as it expands. However, if the gas were returned to its initial state (that is, made to complete a cycle), it would have to be compressed and heat would have to be extracted from it.

The Kelvin statement is a manifestation of a well-known engineering problem. Despite advancing technology, we are not able to build a heat engine that is 100 % efficient. The first law does not exclude the possibility of constructing a perfect engine, but the second law forbids it.

Part a shows schematic of a perfect heat engine with a downward arrow Q at T subscript h and a right arrow W where Q equals W. Part b shows schematic of a perfect refrigerator with an upward arrow Q at T subscript c and an upward arrow Q at T subscript h.
(a) A “perfect heat engine” converts all input heat into work. (b) A “perfect refrigerator” transports heat from a cold reservoir to a hot reservoir without work input. Neither of these devices is achievable in reality.

We can show that the Kelvin statement is equivalent to the Clausius statement if we view the two objects in the Clausius statement as a cold reservoir and a hot reservoir. Thus, the Clausius statement becomes: It is impossible to construct a refrigerator that transfers heat from a cold reservoir to a hot reservoir without aid from an external source . The Clausius statement is related to the everyday observation that heat never flows spontaneously from a cold object to a hot object. Heat transfer in the direction of increasing temperature always requires some energy input . A “ perfect refrigerator ,” shown in [link] (b), which works without such external aid, is impossible to construct.

To prove the equivalence of the Kelvin and Clausius statements, we show that if one statement is false, it necessarily follows that the other statement is also false. Let us first assume that the Clausius statement is false, so that the perfect refrigerator of [link] (b) does exist. The refrigerator removes heat Q from a cold reservoir at a temperature T c and transfers all of it to a hot reservoir at a temperature T h . Now consider a real heat engine working in the same temperature range. It extracts heat Q + Δ Q from the hot reservoir, does work W , and discards heat Q to the cold reservoir. From the first law, these quantities are related by W = ( Q + Δ Q ) Q = Δ Q .

Questions & Answers

The properties of a system during a reversible constant pressure non-flow process at P= 1.6bar, changes from constant volume of 0.3m³/kg at 20°C to a volume of 0.55m³/kg at 260°C. its constant pressure process is 3.205KJ/kg°C Determine: 1. Heat added, Work done, Change in Internal Energy and Change in Enthalpy
Opeyemi Reply
please how do dey get 5/9 in the conversion of Celsius and Fahrenheit
Gwam Reply
what is copper loss
timileyin Reply
this is the energy dissipated(usually in the form of heat energy) in conductors such as wires and coils due to the flow of current against the resistance of the material used in winding the coil.
it is the work done in moving a charge to a point from infinity against electric field
Ashok Reply
what is the weight of the earth in space
peterpaul Reply
As w=mg where m is mass and g is gravitational force... Now if we consider the earth is in gravitational pull of sun we have to use the value of "g" of sun, so we can find the weight of eaeth in sun with reference to sun...
g is not gravitacional forcé, is acceleration of gravity of earth and is assumed constante. the "sun g" can not be constant and you should use Newton gravity forcé. by the way its not the "weight" the physical quantity that matters, is the mass
Yeah got it... Earth and moon have specific value of g... But in case of sun ☀ it is just a huge sphere of gas...
Thats why it can't have a constant value of g ....
not true. you must know Newton gravity Law . even a cloud of gas it has mass thats al matters. and the distsnce from the center of mass of the cloud and the center of the mass of the earth
please why is the first law of thermodynamics greater than the second
Ifeoma Reply
define electric image.obtain expression for electric intensity at any point on earthed conducting infinite plane due to a point charge Q placed at a distance D from it.
Mateshwar Reply
explain the lack of symmetry in the field of the parallel capacitor
Phoebe Reply
pls. explain the lack of symmetry in the field of the parallel capacitor
does your app come with video lessons?
Ahmed Reply
What is vector
Ajibola Reply
Vector is a quantity having a direction as well as magnitude
tell me about charging and discharging of capacitors
Ahemen Reply
a big and a small metal spheres are connected by a wire, which of this has the maximum electric potential on the surface.
Bundi Reply
3 capacitors 2nf,3nf,4nf are connected in parallel... what is the equivalent capacitance...and what is the potential difference across each capacitor if the EMF is 500v
Prince Reply
equivalent capacitance is 9nf nd pd across each capacitor is 500v
four effect of heat on substances
Prince Reply
why we can find a electric mirror image only in a infinite conducting....why not in finite conducting plate..?
Rima Reply
because you can't fit the boundary conditions.
what is the dimensions for VISCOUNSITY (U)
Practice Key Terms 3

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Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
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