The solution to Problem 16 has an error in it.
The specific heat of water is listed as c = 1 J/g K
but should be c = 4.186 J/g K
The final numerical answer given is correct but the math shown does not give that answer.
This print-out should have 22 questions, check that it is complete. Multiple-choice questions may continue on the next column or page:
find all choices before making your selection.
The due time is Central time.
Chapter 22 problems.
001(part 1 of 2) 0 points
A heat engine absorbs 362 J of thermal energy and performs 27.2 J of work in each cycle.
Find the efficiency of the engine.
Correct answer: 0.0751381 . Explanation:
Given : Qh= 362 J and W = 27.2 J
The thermal efficiency of a heat engine is e = W
Qh
= 27.2 J 362 J
= 0.0751381 .
002(part 2 of 2) 0 points
Find the thermal energy expelled in each cy- cle.
Correct answer: 334.8 J.
Explanation:
The work done by a heat engine through a cyclic process (∆U = 0) is
W = Qh−Qc
Qc = Qh−W
= 334.8 J .
003(part 1 of 3) 4 points
Determine the change in the internal energy of a system that absorbs 555 cal of thermal energy while doing 580 J of external work.
Correct answer: 1743.23 J.
Explanation:
Given : Q = 555 cal and W = 580 J .
According to the first law of thermodynamics, we have
∆U = Q − W
where Q is the thermal energy transferred into the system and W is the work done by the system. Then we have
∆U = 555 cal 4.186 J
cal −580 J
= 1743.23 J .
004(part 2 of 3) 3 points
Determine the change in the internal energy of a system that absorbs 769 cal of thermal energy while 626 J of external work is done on the system.
Correct answer: 3845.03 J.
Explanation:
∆U = Q − (−W )
= 769 cal 4.186 J
cal + 626 J
= 3845.03 J .
005(part 3 of 3) 3 points
Determine the change in the internal energy of a system that is maintained at a constant volume while 1270 cal is removed from the system.
Correct answer: −5316.22 J.
Explanation:
Since the volume is maintained constant W = P ∆V = 0
∆U = Q
= −1270 cal 4.186 J cal
= −5316.22 J .
006(part 1 of 2) 0 points
An ideal gas is compressed to half its original volume while its temperature is held constant.
If 750 J of energy is removed from the gas during the compression, how much work is done on the gas?
Correct answer: 750 J.
Explanation:
Given : Q = 750 J .
According to the first law of thermodynamics,
∆U = Q − W ,
where Q is the thermal energy transferred into the system and W is the work done by the system. Since ∆U = 0, then
W = Q
= 750 J .
007(part 2 of 2) 0 points
What is the change in the internal energy of the gas during the compression?
Correct answer: 0 J.
Explanation:
If the temperature remains constant,
∆U = 0 J .
008(part 1 of 2) 5 points
A Carnot engine has a power output of 126 kW. The engine operates between two reservoirs at 16◦C and 655◦C.
How much thermal energy is absorbed per hour?
Correct answer: 6.58749 × 108 J.
Explanation:
Given : P = 126 kW ,
Th= 655◦C = 928 K , and Tc= 16◦C = 289 K .
The efficiency of heat engine is eff = W
Qh
= 1 − Tc
Th
and the work done by the system is W = P t .
Thus
Qh= W eff
= P t 1 − Tc
Th
= (126 kW) (3600 s) 1 − 289 K
928 K
= 6.58749 × 108 J .
009(part 2 of 2) 5 points
How much thermal energy is lost per hour?
Correct answer: 2.05149 × 108 J.
Explanation:
The work done by a heat engine through a cyclic process (∆U = 0) is
W = Qh−Qc. Then
Qc= Qh−W
= Qh−P t
= (6.58749 × 108 J)
−(126 kW) (3600 s) W KW
= 2.05149 × 108 J .
010(part 1 of 2) 0 points
A steam engine is operated in a cold climate where the exhaust temperature is −26◦C.
Calculate the theoretical maximum effi- ciency of the engine using an intake steam temperature of 114◦C.
Correct answer: 0.361757 . Explanation:
Given : Th = 114◦C = 387 K and Tc = −26◦C = 247 K .
According to Carnot’s theorem, the theoreti- cal maximum efficiency is
e = 1 − Tc
Th
= 1 − 247 K 387 K
= 0.361757 .
011(part 2 of 2) 0 points
If, instead, superheated steam at 286◦C is used, find the maximum possible efficiency.
Correct answer: 0.55814 . Explanation:
Th = 286◦C = 559 K The maximum efficiency is
e = 1 − 247 K 559 K
= 0.55814 .
012(part 1 of 1) 0 points
The efficiency of a 840 MW nuclear power plant is 27.2 %.
If a river of flow rate 3.84 × 106 kg/s were used to transport the excess thermal energy away, what would be the average temperature increase of the river?
Correct answer: 0.1394◦C.
Explanation:
Given : Poutput = 840 MW = 106W and e = 27.2 % = 0.272 .
The excess thermal energy transported per second by the river is
Pexcess = Pinput(1 − e)
=µ Poutput
e
¶
(1 − e)
=µ 840 MW 0.272
¶
(1 − 0.272)
= 2248.24 MW
where e is efficiency and Poutput is power out- put of the plant. Then the temperature of the river is increased (per second) by
dm
dt c ∆T = dQ
dt = Pexcess
where c is heat capacity of water and dm dt is flow rate of the water. Thus
∆T = Pexcess
3.84 × 106 kg/s
= 0.1394◦C .
013(part 1 of 2) 0 points
A house loses thermal energy through the exterior walls and roof at a rate of 4860 W when the interior temperature is 20.1◦C and the outside temperature is −0.2◦C.
Calculate the electric power required to maintain the interior temperature at Ti for the following two cases: The electric power is used in electric resistance heaters (which con- vert all of the electricity supplied to thermal energy).
Correct answer: 4860 W.
Explanation:
Given : ∆Q/∆t = 4860 W , Ti= 20.1◦C , and To= −0.2◦C .
Since all the electricity supplied is converted to thermal energy, we have
∆Q
∆t = ∆E
∆t = PEl
Thus
PEl = 4860 W .
014(part 2 of 2) 0 points
The electric power is used to operate the com- pressor of a heat pump (which has a coefficient of performance equal to ν = 0.7 of the Carnot cycle value).
Correct answer: 480.86 W.
Explanation:
For a heat pump we have (COP )Carnot= Ti
Ti−To
= 20.1◦C + 273 K 20.1◦C − (−0.2◦C)
= 14.4384
Hence to bring 4860 W of heat in the house requires only
Ph = ∆Q/∆t (COP )actual
= W
0.6 (COP )carnot
= 4860 W (0.7) (14.4384)
= 480.86 W .
015(part 1 of 1) 0 points
An ice tray contains 375 g of water at 0◦C.
Calculate the change in entropy of the wa- ter as it freezes completely and slowly at 0◦C.
Correct answer: −457.418 J/K.
Explanation:
Given : m = 375 g = 0.375 kg , L = 333000 J/kg , and T = 0◦C = 273 K . In the freezing process T is constant, so
∆Q = −m L
where m is mass of water, and l is latent heat of fusion. Thus
∆S = ∆Q
T = −m L T
= −(0.375 kg)(333000 J/kg) 273 K
= −457.418 J/K .
016(part 1 of 1) 10 points
Calculate the change in entropy of 210 g of water heated slowly from 15.8◦C to 80.5◦C.
Correct answer: 177.701 J/K.
Explanation:
Given : Ti= 15.8◦C = 288.8 K , Tf = 80.5◦C = 353.5 K ,
m = 210 g , and c = 1 J/g · K . The heat absorbed in the process is
dQr = m c dT .
The change in entropy in an arbitrary re- versible process between an initial state and final state is
∆S = Z f
i
dS = Z f
i
dQr
T
= Z f
i
m cdT T
= m c logTf
Ti
= (210 g) (1 J/g · K) logµ 353.5 K 288.8 K
¶
= 177.701 J/K .
017(part 1 of 6) 2 points
One mole of an ideal monatomic gas is taken through the cycle “abca” shown schematically in the diagram. State “a” has volume Va = 0.0159 m3 and pressure Pa = 121000 Pa, and state “c” has volume Vc= 0.0526 m3. Process
“ca” lies along the T = 231 ±1 K isotherm.
The molar heat capacities for the gas are cp = 20.8 J/mol · K and cv = 12.5 J/mol · K.
0 1.2 0.9 0.6 0.3 1.5
0 17 34 51 68 85 p (×105Pa)
V (×10-3m3)
a b
c 250 K
250 K
This schematic plot is intended to give an example of a P V diagram (not to scale). Use the values of P , V , and T given above.
Determine the temperature Tbof state “b”.
Correct answer: 765.485 K.
Explanation:
Given :
Pb = 121000 Pa , Vb = 0.0526 m3, and Tb = 8.31447 J/mol · K.
We use the ideal gas equation T = P V
n R ,
where P is the pressure, V is the volume (both evaluated at “b”), R is the molar gas constant, and n is the number of moles.
Tb = P V R
= (121000 Pa) (0.0526 m3) 8.31447 J/mol · K
= 765.485 K .
018(part 2 of 6) 2 points
Determine the heat Qabadded to the gas dur- ing process “ab”.
Correct answer: 11109.1 J.
Explanation:
Given :
Pa = 121000 Pa , Va = 0.0159 m3, and Ta = 8.31447 J/mol · K.
For state “a”
Ta = P V R
= (121000 Pa) (0.0159 m3) 8.31447 J/mol · K
= 231.392 K
Thus
Q = n cp∆T
= (1 mol)(20.8 J/mol · K)
×(765.485 K − 231.392 K)
= 11109.1 J ,
where Q is the heat transferred, n is the num- ber of moles, cpis the the molar heat capacity for a constant pressure process (such as pro- cess “ab”), and ∆T is the change in tempera- ture from “a” to “b”.
019(part 3 of 6) 2 points
Determine the change in the internal energy
∆Uab = Ub−Ua.
Correct answer: 6668.43 J.
Explanation:
In an isobaric process the change in internal energy is given by
∆Uab = Qab−W
= Qab−P ∆V
= Qab−P [Vb−Va]
= 11109.1 J − (121000 Pa)
×(0.0526 m3−0.0159 m3)
= 6668.43 J ,
020(part 4 of 6) 2 points
Determine the work Wbc done by the gas on its surroundings during process “bc”.
Correct answer: 0 . Explanation:
W = P ∆V and ∆V = 0, so W = 0 . 021(part 5 of 6) 1 points
The net heat added to the gas for the entire cycle is 2140 J.
Determine the net work done by the gas on its surroundings for the entire cycle.
Correct answer: 2140 J.
Explanation:
Given : Q = 2140 J .
For a complete cycle the change in internal energy ∆U is zero, so
W = Q = 2140 J .
The work is simply the net heat added to the gas.
022(part 6 of 6) 1 points
Determine the efficiency Eff of a Carnot en- gine that operates between the maximum and minimum temperatures in this cycle.
Correct answer: 0.697719 . Explanation:
The Carnot efficiency Eff is given by Eff = 1 − Tc
Th
.
The maximum temperature is clearly that of state “b”, determined to be 765.485 K in ques- tion 1. The minimum temperature will be that of the isotherm, 231.392 K. Therefore
Eff = 1 − Ta
Tb
= 1 − 231.392 K 765.485 K
= 0.697719 .
