Soluções Listas Termodinâmica

Prévia do material em texto

1 Lista 1
Problema 2) E…ciência de um refrigerador de Carnot:
k =
QF
W
=
QF
jQQj �QF =
1
jQQj
QF
� 1
onde
W = jQQj �QF
k =
QF
W
=
QF
jQQj �QF =
1
jQQj
QF
� 1
Usando a relação válida para o ciclo de Carnot
jQQj
QF
=
TQ
TF
obtemos
k =
1
TQ
TF
� 1
Cálculo do calor:
Q = mc�T +mL
Q = 10� 2:05� 103 � 20 + 10� 340� 103
Q = 3:82� 106J
TQ = 303; 15 K; TF = 253; 15 K
E…ciência:
k =
1
303:15
253:15 � 1
= 5:06
Por de…nição
k =
QF
W
e
W =
QF
k
Obtemos
1
W = 7:549 4� 105 J
W = 754:9 4 kJ =
754:9 4
3600
kWh = 0:21kWh
Problema 3)
P = 1200 W
TQ = 313; 15 K; TF = 293; 15 K
Então
k =
1
313:15
293:15 � 1
k = 14:65
jQQj = QF +W = (k + 1)P�t
jQQj = 15:65� 1200� 3600
jQQj = 6:76� 107 J
Problema 5)
Q = ncp�T ! �T = Q
ncp
�V =
nR
p
�T =
nR
p
Q
ncp
=
RQ
5
2Rp
=
2Q
5p
W = p�V = p
2Q
5p
W =
2
5
Q!W = 200 J
Problema 6)
�U = 0!W = Q
W = Q = T�S = 400� 50 = 20 kJ
Problema 8)
(a)
2
pbVb = paVa
pb = pa
Va
Vb
! pb = 0; 5 atm
pcV
c = paV
a
pc = pa
�
Va
Vc
�
! pc = (0; 5)1;4 = 0; 379 atm
(b)
Qab = paVa ln
Vb
Va
= 1; 0 ln 2:0 = 0; 693 atm:l
Qbc = �U = ncvTc � ncvTb
Qbc =
cv
R
(pcVc � pbVb)
Qbc =
5
2
(0; 379� 2; 00� 0; 500� 2; 00) = �0:605 atm:l
� = 1� jQbcj
Qab
� = 1:0� 0:605
0:693
= 0:127
(c)
W = Qab � jQbcj
W = 0:693� 0:605 = 0:088 atm:l
W = 0:088� 100 = 8; 8 J
Problema 10)
(a)
QAB =
cp
R
pA (VB � VA) = 5
2
� 5; 0 = 12; 5 atm:l
QCD =
cp
R
pC (VD � VC) = 5
2
� 0; 5 (15; 2� 22; 8) = �9; 5 atm:l
3
Rendimento
� = 1� jQCDj
QAB
� = 1; 0� 9; 5
12; 5
= 0; 24
(b) Invertendo o ciclo, a e…ciência do refrigerador …ca dada por
k =
QCD
W
=
QCD
jQAB j �QCD
k =
9; 5
12; 5� 9; 5 ! k = 3; 2
Problema 12)
(a)
Q12 = �U12 = ncv (T2 � T1)
Q12 =
1
2
� 3
2
R (1000� 250)
Q12 =
3
4
� 8; 314 (1000� 250) = 4:676; 6 J
Q23 =
Z S3
S2
T2ds = T2 (S3 � S2)
Q23 = 1000 (20� 28) = �8:000 J
k =
Q12
jQ23j �Q12
k =
4:676; 6
8:000� 4:676; 6 = 1; 4
(b) Veja a …gura do problema 8.
2 Lista 2
Problema 2)
W1 = Q1 �Q2;
W2 = Q2 �Q3
� =
W
Q1
=
W1 +W2
Q1
=
Q1 �Q3
Q1
4
� = 1� Q3
Q1
Q1
Q2
=
T1
T2
;
Q3
Q2
=
T3
T2
Escrevendo
Q2 = Q1
T2
T1
obtemos
Q3
Q2
=
Q3
Q1
T1
T2
=
T3
T2
! Q3
Q1
=
T3
T1
Portanto, demonstra-se que
� = 1� T3
T1
Problema 3)
(a)
k =
Qf
W
; W = jQqj �Qf
jQqj = Qf +W = Qf + Qf
k
jQqj = 42 + 42
5; 7
= 49; 37 kcal
(b)
W =
42
5; 7
= 7; 37 kcal
Problema 4)
(a)
Q = 500 cal!W = Q = 500 cal
(b) Rendimento da máquina:
� = 1� jQf j
Qq
= 1� T2
T1
� = 1� T2
T1
= 1� 300
400
= 0; 25
Logo,
jQf j = Qq T2
T1
jQf j = 500� 300
400
= 375 cal
(c)
W = jQf j = 375 cal
5
Problema 6)
Q = mcg�T +mLF +mc�T
0
Qg = 10� 0; 50� 10 + 10� 80 + 10� 1; 0� 15
Qg = 1; 0 kcal
�S =
Qg
Tlago
=
1000:0
288:
= 3; 47 cal=K
Problema 7)
(a)
QI = nRT ln
V 0
V
+ ncv (T"� T 0)
QI = pV ln
V 0
V
+ cv
�
p"V 0
R
� p
0V 0
R
�
QI = pV ln
V 0
V
+
cv
R
(p"� p0)V 0
QI = pV ln 2 +
cv
R
�
2p� p
2
�
2V = pV ln 2 +
3cv
R
V
QI = pV ln 2 + 4; 5pV
p0 = 2p; V 0 = V=2; p" = 2p; V " = 2V
QII = pV ln
V 0
V
+
cp
R
p0 (V "� V 0) = pV ln V
0
V
+
2cp
R
p
�
2V � V
2
�
QII = pV ln
V 0
V
+
3cp
R
pV = �pV ln 2 + 3cp
R
pV
QII = �pV ln 2 + 7; 5pV
(b)
WI = pV ln
V 0
V
= pV ln 2
WII = pV ln
V 0
V
+ 2p (V "� V 0) = �pV ln 2 + 3pV
(c)
6
�UI = ncv (T"� T 0) = cv
R
�
2p� p
2
�
V 0 =
cv
R
3
2
p (2V )
�UI =
3cv
R
pV = 4; 5pV
�UII = 2p�V = 2p
3
2
V = 3pV
(d)
�SI =
nRT ln V
0
V
T
+
Z T"
T 0
dQ
T
= R ln
V 0
V
+
Z T"
T 0
ncvdT
T
�SI = R ln 2 + cv ln
T"
T 0
�SI = R ln 2 + cv ln
p"V "
p0V 0
�SI = R ln 2 + cv ln
2p
p
2
= R ln 2 + 2cv ln 2
�SI = 4R ln 2
�SII =
nRT ln V
0
V
T
+
Z T"
T 0
dQ
T
= R ln
V 0
V
+
Z T"
T 0
ncpdT
T
�SII = �R ln 2 + cp ln T"
T 0
�SII = �R ln 2 + cp ln p"V "
p0V 0
�SII = �R ln 2 + cp ln 2VV
2
= �R ln 2 + 2cp ln 2
�SII = 4R ln 2
Problema 9)
(a)
p = 5; 00� 103eVi�Va N
m2
7
pf = 5; 00� 103e
Vi�Vf
a = 5; 00� 103e�1
pf = 1:839; 4
N
m2
(b)
pfVf = nRTf
Tf =
pfVf
R
=
1:839; 4� 2; 00
8; 314
Tf = 442 K
(c)
W =
Z Vf
Vi
pdV
W = 5; 00� 103
Z Vf
Vi
e
Vi�V
a dVZ 2
1
e1�xdx = 0; 632
W = 5; 00� 103 � 0; 632
W = 3:160 J
(d) Processo isobárico:
�S1 =
Z Tf
Ti
dQ
T
=
Z Tf
Ti
ncpdT
T
= ncp ln
Tf
Ti
= ncp ln
pfVf
piVi
�S1 = cp ln
Vf
Vi
= 2; 5R ln 2
Processo isovolumétrico:
�S2 =
Z Tf
Ti
dQ
T
=
Z Tf
Ti
ncvdT
T
= ncv ln
Tf
Ti
= ncv ln
pfVf
piVi
�S2 = cv ln
pf
pi
= 1; 5R ln
1:839; 4
5:000
�S2 = �1; 5R
�S = �S1 +�S2 = 2; 5R ln 2� 1; 5R
�S = 0; 233R
Processo isotérmico:
�S1 =
Q
Ti
=
nRTi ln
Vf
Vi
Ti
= nR ln
Vf
Vi
= R ln 2
8
Processo isovolumétrico:
�S2 =
Z Tf
Ti
dQ
T
=
Z Tf
Ti
ncvdT
T
= ncv ln
Tf
Ti
= ncv ln
pfVf
piVi
�S2 = cv ln
pf
pi
= 1; 5R ln
1:839; 4
5:000
piVf = nRTi = p0V0
pi =
p0V0
Vf
=
5:000
2; 0
= 2:500
�S2 = 1; 5R ln
1:839; 4
2:500
�S2 = �0:460 28R
�S = R ln 2� 0:460 28R
�S = 0; 233R
Problema 10)
k =
Qf
W
;
d
dt
jQqj = 1; 8� 106 cal=hora
W = jQqj �Qf = jQqj � kW
W =
jQqj
1 + k
dW
dt
=
1
1 + k
d
dt
jQqj
dW
dt
=
1
1 + k
1; 8� 106
3:600
cal
s
dW
dt
=
500
4; 8
cal
s
= 104; 17
cal
s
dW
dt
= 104; 17� 4; 186 cal
s
dW
dt
= 436 W
9