solutions of thermodynamics prs

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27 Οκτ 2013 (πριν από 3 χρόνια και 9 μήνες)

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RESULTS OF THERMAL PROBLEMS
28.10.13

p #
1

RESULTS OF THERMODYNAMICS PROBLEMS



PROBLEM # 1:


a) c = 1000 J/kg
-
o
C

b) L = 50000 J/kg

c) (i) AB and CD

(ii) AB; BC and CD


PROBLEM # 2:


a) W = 100t

b) t = 33.4 s

c) t = 461 s


PROBLEM # 3:


a)

b) t = 1000 s












PROBLEM # 4:


a
)

b)

c)


d)

PROBLEM # 5:


a) p
B

= p
0
/2

b) W =
-

1000 J

c) Since Q

0 then

S

0

d) Q
c

=
-
400 J


e) W
T

=
-
600 J

done by gas



PROBLEM # 6:

a)


b) (i)

W =
-

2 p
0

V
0

(ii)



u = 0






(iii) Q
T

= 2 p
0

V
0






c) Q
II

= 10 p
0

V
0










PROBLEM # 7:

a) P = 400 J

b) e = 0.4

c) (i) Q
H

= 1000 J/s

(ii) Q
C

= 600 J/s


PROBLEM # 8:

a) p = 8.2 x 10
4

pa

b) F = 41
0

N

c)

n = 0.0055 mol.


PROBLEM # 9:

a) T
A

= 300 K, T
B

=
6
00 K , T
C

= 600 K

b) W =
-
1250 J

c) Q
T
= 1250 J


d) Q =
-
6240 J

e) e = 0.17



PROBLEM # 10:

a) P = 300 MW

b) 6.8 kg/s

c) Q = 180 MW

d)

T = 1.55
0
C


PROBLEM # 11:

a) I = 1.5 A

b) P = 90 W

c) c = 18 kJ/kg
-
0
C

d) t = 30 s



e)












V

A

R
1

R
2

p


V

3V
o

2V
o

V
o

2p
o

p
o

A

B

C

D

I

II

III

IV

T
0

2T
0

6T
0

3T
0

t

20

40

60

50

100

T (
o
C)

80

100

200

150

250

300

RESULTS OF THERMAL PROBLEMS
28.10.13

p #
2

PROBLEM # 12:

a) T
b

= 750 K

b) Q = 10400 J

c)

u = 6240 J

d) W = 0


e) W =
-

3300 J

f) e = 0.66


PROBLEM # 13:

a) (i) P =
6.2

W (ii) P = 57.6 W b)



c) (i) t = 2734 s (ii) longer



d) To breake the liquid stractu
re









PROBLEM # 14:

a) W = Mgh

b)

T =gh/c

c)

T = 15.7
0
C

d) No e) Heat lost


PROBLEM # 15:

a) p
c

= 2x10
5

pa

b)



c)

W


0
done by gas




d) Heat Engine since Q
H


Q
C














PROBLEM # 16:


a)

b) T
i
, T
f

, I, V, t, m

c) Q
Taken

=

m c

T, P = I V , Q
Given

= P t , Q
T

= Q
G




d) Neglecting the heat taken by the calorimeter we will


get larger c .









calorimeter


heating coil


ammeter


voltmeter


cables


thermometer


power supplier


sample liqui
d








PROBLEM # 17:


Follow

the lecture notes


PROBLEM # 18:


a) p
2

= 1.04 x10
5

pa

b) V
2

= 1.47 x 10
-
3

m
3


c)

p = 0 then isobaric

d)
Isobaric


e)
V
4

= 2.05 x 10
-
3

m
3



PROBLEM # 19:


a)p = 1.5 x 10
5

pa

b) p = 1.37 x 10
5

pa

c) m = 0.44 kg


PROBLEM

# 20:


a)

u = 563 J

b) i) removed ii) Q =
-
640 J

c) W= 450 J d) added
,

+

-

battery

V

A

R

R

R

0

2

4

6

8

1

3

2

V (x 10
-
2

m
3
)

4

p (x 10
5

pa)

B

A

C

RESULTS OF THERMAL PROBLEMS
28.10.13

p #
3


PROBLEM # 21:


a) T
2
= T
1
/2

b) P
2

= 2 P
1


c) W
ABC
= P
1
P
2
/2

d)
BCA




PROBLEM # 22:


a) (i) 3600J (ii)

u = 5400 J (iii) Q = 9000 J


b)


c)

(i) (ii) Q < 0 given








PROBLEM # 23:


a) H = (nR/pA) T

b)








c) 1.136 moles










PROBLEM # 24:


a) p = nRT (1/V)

b)









c)












d)

n = 2.4 x 10
-
3

moles



PROBLEM # 25:


a) (i) T
2

= 746 K (ii)

T
3
= 1119 K

b) W = 12500 J done on the gas

c) Q= 12500 J removed,





m (kg)

V

(m
3
)

1/V (m
-
3
)

P (pa)

0

6.0 x10
-
5

1.7 x10
4

1.02x10
5

1

4.5 x10
-
5

2.2 x10
4

1.35x10
5

2

3.6 x10
-
5

2.8 x10
4

1.68x10
5

3

3.0 x10
-
5

3.3 x10
4

2.01x10
5

4

2.6 x10
-
5

3.8 x10
4

2.34x10
5

P
(N/m
2
)

0

2
0
0

4
0
0

6
0
0

8
0
0

0

5

10

15

20

B


A


V(
m
3
)

C


H (m)

T (K)

1.1

1.
3

1.
5

300

350

400

450

H (m)

T (K)

1.1

1.
3

1.
5

300

350

400

450

P(10
5
pa)

2

1

3

2

3

4

1/V(10
4

m
-
3
)