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A 0.50-kg piece of aluminum at 250 ° C is dropped into 1.0 kg of water at 20 ° C . After equilibrium is reached, what is the net entropy change of the system?

82 J/K

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Suppose 20 g of ice at 0 ° C is added to 300 g of water at 60 ° C . What is the total change in entropy of the mixture after it reaches thermal equilibrium?

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A heat engine operates between two temperatures such that the working substance of the engine absorbs 5000 J of heat from the high-temperature bath and rejects 3000 J to the low-temperature bath. The rest of the energy is converted into mechanical energy of the turbine. Find (a) the amount of work produced by the engine and (b) the efficiency of the engine.

a. 2000 J; b. 40 %

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A thermal engine produces 4 MJ of electrical energy while operating between two thermal baths of different temperatures. The working substance of the engine rejects 5 MJ of heat to the cold temperature bath. What is the efficiency of the engine?

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A coal power plant consumes 100,000 kg of coal per hour and produces 500 MW of power. If the heat of combustion of coal is 30 MJ/kg, what is the efficiency of the power plant?

60 %

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A Carnot engine operates in a Carnot cycle between a heat source at 550 ° C and a heat sink at 20 ° C . Find the efficiency of the Carnot engine.

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A Carnot engine working between two heat baths of temperatures 600 K and 273 K completes each cycle in 5 sec. In each cycle, the engine absorbs 10 kJ of heat. Find the power of the engine.

64.4 %

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A Carnot cycle working between 100 ° C and 30 ° C is used to drive a refrigerator between −10 ° C and 30 ° C . How much energy must the Carnot engine produce per second so that the refrigerator is able to discard 10 J of energy per second?

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Challenge problems

(a) An infinitesimal amount of heat is added reversibly to a system. By combining the first and second laws, show that d U = T d S d W . (b) When heat is added to an ideal gas, its temperature and volume change from T 1 and V 1 to T 2 and V 2 . Show that the entropy change of n moles of the gas is given by

Δ S = n C v ln T 2 T 1 + n R ln V 2 V 1 .

derive

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Using the result of the preceding problem, show that for an ideal gas undergoing an adiabatic process, T V γ 1 is constant.

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With the help of the two preceding problems, show that Δ S between states 1 and 2 of n moles an ideal gas is given by

Δ S = n C p ln T 2 T 1 n R ln p 2 p 1 .

derive

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A cylinder contains 500 g of helium at 120 atm and 20 ° C . The valve is leaky, and all the gas slowly escapes isothermally into the atmosphere. Use the results of the preceding problem to determine the resulting change in entropy of the universe.

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A diatomic ideal gas is brought from an initial equilibrium state at p 1 = 0.50 atm and T 1 = 300 K to a final stage with p 2 = 0.20 atm and T 1 = 500 K . Use the results of the previous problem to determine the entropy change per mole of the gas.

18 J/K

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The gasoline internal combustion engine operates in a cycle consisting of six parts. Four of these parts involve, among other things, friction, heat exchange through finite temperature differences, and accelerations of the piston; it is irreversible. Nevertheless, it is represented by the ideal reversible Otto cycle , which is illustrated below. The working substance of the cycle is assumed to be air. The six steps of the Otto cycle are as follows:

  1. Isobaric intake stroke ( OA ). A mixture of gasoline and air is drawn into the combustion chamber at atmospheric pressure p 0 as the piston expands, increasing the volume of the cylinder from zero to V A .
  2. Adiabatic compression stroke ( AB ). The temperature of the mixture rises as the piston compresses it adiabatically from a volume V A to V B .
  3. Ignition at constant volume ( BC ). The mixture is ignited by a spark. The combustion happens so fast that there is essentially no motion of the piston. During this process, the added heat Q 1 causes the pressure to increase from p B to p C at the constant volume V B ( = V C ) .
  4. Adiabatic expansion ( CD ). The heated mixture of gasoline and air expands against the piston, increasing the volume from V C to V D . This is called the power stroke , as it is the part of the cycle that delivers most of the power to the crankshaft.
  5. Constant-volume exhaust ( DA ). When the exhaust valve opens, some of the combustion products escape. There is almost no movement of the piston during this part of the cycle, so the volume remains constant at V A ( = V D ) . Most of the available energy is lost here, as represented by the heat exhaust Q 2 .
  6. Isobaric compression ( AO ). The exhaust valve remains open, and the compression from V A to zero drives out the remaining combustion products.

(a) Using ( i ) e = W / Q 1 ; ( ii ) W = Q 1 Q 2 ; and ( iii ) Q 1 = n C v ( T C T B ) , Q 2 = n C v ( T D T A ) , show that

e = 1 T D T A T C T B .

(b) Use the fact that steps (ii) and (iv) are adiabatic to show that

e = 1 1 r γ 1 ,

where r = V A / V B . The quantity r is called the compression ratio of the engine.

(c) In practice, r is kept less than around 7. For larger values, the gasoline-air mixture is compressed to temperatures so high that it explodes before the finely timed spark is delivered. This preignition causes engine knock and loss of power. Show that for r = 6 and γ = 1.4 (the value for air), e = 0.51 , or an efficiency of 51 % . Because of the many irreversible processes, an actual internal combustion engine has an efficiency much less than this ideal value. A typical efficiency for a tuned engine is about 25 % to 30 % .

The figure shows a closed loop graph with four points A, B, C and D. The x-axis is V and y-axis is p. The value of V at A and D is equal and at B and C is equal.
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An ideal diesel cycle is shown below. This cycle consists of five strokes. In this case, only air is drawn into the chamber during the intake stroke OA . The air is then compressed adiabatically from state A to state B , raising its temperature high enough so that when fuel is added during the power stroke BC , it ignites. After ignition ends at C , there is a further adiabatic power stroke CD . Finally, there is an exhaust at constant volume as the pressure drops from p D to p A , followed by a further exhaust when the piston compresses the chamber volume to zero.

(a) Use W = Q 1 Q 2 , Q 1 = n C p ( T C T B ) , and Q 2 = n C v ( T D T A ) to show that e = W Q 1 = 1 T D T A γ ( T C T B ) .

(b) Use the fact that A B and C D are adiabatic to show that

e = 1 1 γ ( V C V D ) γ ( V B V A ) γ ( V C V D ) ( V B V A ) .

(c) Since there is no preignition (remember, the chamber does not contain any fuel during the compression), the compression ratio can be larger than that for a gasoline engine. Typically, V A / V B = 15 and V D / V C = 5 . For these values and γ = 1.4 , show that ε = 0.56 , or an efficiency of 56 % . Diesel engines actually operate at an efficiency of about 30 % to 35 % compared with 25 % to 30 % for gasoline engines.

The figure shows a closed loop graph with four points A, B, C and D. The x-axis is V and y-axis is p. The value of V at A and D is equal and the value of p at B and C is equal.

proof

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Consider an ideal gas Joule cycle, also called the Brayton cycle, shown below. Find the formula for efficiency of the engine using this cycle in terms of P 1 , P 2 , and γ .

The figure shows a closed loop graph with four points 1, 2, 3 and 4. The x-axis is V and y-axis is p. The value of p at 1 and 4 is equal and at 2 and 3 is equal
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Derive a formula for the coefficient of performance of a refrigerator using an ideal gas as a working substance operating in the cycle shown below in terms of the properties of the three states labeled 1, 2, and 3.

The figure shows a closed loop graph with three points 1, 2 and 3. The x-axis is V and y-axis is p. The value of V at 1 and 2 is equal and the value of p at 2 and 3 is equal.

K R = 3 ( p 1 p 2 ) V 1 5 p 2 V 3 3 p 1 V 1 p 2 V 1

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Two moles of nitrogen gas, with γ = 7 / 5 for ideal diatomic gases, occupies a volume of 10 −2 m 3 in an insulated cylinder at temperature 300 K. The gas is adiabatically and reversibly compressed to a volume of 5 L. The piston of the cylinder is locked in its place, and the insulation around the cylinder is removed. The heat-conducting cylinder is then placed in a 300-K bath. Heat from the compressed gas leaves the gas, and the temperature of the gas becomes 300 K again. The gas is then slowly expanded at the fixed temperature 300 K until the volume of the gas becomes 10 −2 m 3 , thus making a complete cycle for the gas. For the entire cycle, calculate (a) the work done by the gas, (b) the heat into or out of the gas, (c) the change in the internal energy of the gas, and (d) the change in entropy of the gas.

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A Carnot refrigerator, working between 0 ° C and 30 ° C is used to cool a bucket of water containing 10 −2 m 3 of water at 30 ° C to 5 ° C in 2 hours. Find the total amount of work needed.

W = 110,000 J

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Questions & Answers

a body is projected vertically upward of 30kmp/h how long will it take to reach a point 0.5km bellow e point of projection
Abu Reply
i have to say. who cares. lol. why know that t all
Jeff
is this just a chat app about the openstax book?
Lord Reply
kya ye b.sc ka hai agar haa to konsa part
MPL Reply
what is charge quantization
Mayowa Reply
it means that the total charge of a body will always be the integral multiples of basic unit charge ( e ) q = ne n : no of electrons or protons e : basic unit charge 1e = 1.602×10^-19
Riya
is the time quantized ? how ?
Mehmet
What do you meanby the statement,"Is the time quantized"
Mayowa
Can you give an explanation.
Mayowa
there are some comment on the time -quantized..
Mehmet
time is integer of the planck time, discrete..
Mehmet
planck time is travel in planck lenght of light..
Mehmet
it's says that charges does not occur in continuous form rather they are integral multiple of the elementary charge of an electron.
Tamoghna
it is just like bohr's theory. Which was angular momentum of electron is intral multiple of h/2π
Aditya
determine absolute zero
OFERE Reply
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
U can easily calculate work done by 2.303log(v2/v1)
Abhishek
Amount of heat added through q=ncv^delta t
Abhishek
Change in internal energy through q=Q-w
Abhishek
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.
Henry
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...
Prince
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
Jorge
Yeah got it... Earth and moon have specific value of g... But in case of sun ☀ it is just a huge sphere of gas...
Prince
Thats why it can't have a constant value of g ....
Prince
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
Jorge
please why is the first law of thermodynamics greater than the second
Ifeoma Reply
every law is important, but first law is conservation of energy, this state is the basic in physics, in this case first law is more important than other laws..
Mehmet
First Law describes o energy is changed from one form to another but not destroyed, but that second Law talk about entropy of a system increasing gradually
Mayowa
first law describes not destroyer energy to changed the form, but second law describes the fluid drection that is entropy. in this case first law is more basic accorging to me...
Mehmet
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
Phoebe
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
Damilare
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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