Exams Net
Unrestricted Exams Practice
Home
Exams
Banking
CUET
Defence
Engineering
Finance
GATE
Insurance
International
JEE
LAW
MBA
MCA
Medical
Other
Police
PSC
RRB
SSC
State Govt
Subjectwise
Teacher
SET Exams
UPSC
Aptitude
Algebra and Higher Mathematics
Arithmetic
Commercial Mathematics
Data Based Mathematics
Geometry and Mensuration
Number System and Numeracy
Problem Solving
Board Exams
Andhra
Bihar
CBSE
Gujarat
Haryana
ICSE
Jammu and Kashmir
Karnataka
Kerala
Madhya Pradesh
Maharashtra
Odisha
Tamil Nadu
Telangana
Uttar Pradesh
English
Competitive English
CBSE
CBSE Class 10 Solutions
CBSE Class 12 Solutions
CBSE Question Papers (Pdf)
NCERT Books (Pdf)
NCERT Exemplar Books (Pdf)
NCERT Study Notes (Pdf)
CBSE Study Concepts (Pdf)
NCERT Text Book Class 11 Solutions
NCERT Text Book Class 12 Solutions
ICSE Class 10 Papers
Certifications
Technical
Cloud Tech Certifications
Security Tech Certifications
Management
IT Infrastructure
More
About
Contact Us
Our Apps
Privacy
+
-
Test Index
Thermodynamics Part 4
Show Para
Hide Para
Section:
Physics
Share question:
© examsnet.com
Question : 57 of 100
Marks:
+1
,
-0
1
kg
1 \text{kg}
1
kg
of water at
10
0
∘
C
100^{\circ} \text{C}
10
0
∘
C
is converted into steam at
10
0
∘
C
100^{\circ} \text{C}
10
0
∘
C
by boiling at atmospheric pressure. The volume of water changes from
1.00
×
1
0
−
3
m
3
1.00 \times 10^{-3} \text{m}^3
1.00
×
1
0
−
3
m
3
as a liquid to
1.671
m
3
1.671 \text{m}^3
1.671
m
3
as steam. The change in internal energy of the system during the process will be
(Given latent heat of vaporisation
=
2257
kJ
/
kg
=2257 \text{kJ}/\text{kg}
=
2257
kJ
/
kg
, Atmospheric pressure
=
1
×
1
0
5
Pa
=1 \times 10^5 \text{Pa}
=
1
×
1
0
5
Pa
)
[11-Apr-2023 shift 1]
+
2476
kJ
+2476 \text{kJ}
+
2476
kJ
−
2426
kJ
-2426 \text{kJ}
−
2426
kJ
−
2090
kJ
-2090 \text{kJ}
−
2090
kJ
+
2090
kJ
+2090 \text{kJ}
+
2090
kJ
Validate
Solution:
Change in volume at constant pressure and temp
→
\rightarrow
→
Δ
V
=
V
2
−
V
1
=
1.671
−
0.001
\Delta V = \; V_2- V_1 = 1.671-0.001
Δ
V
=
V
2
−
V
1
=
1.671
−
0.001
Δ
V
=
1.67
m
3
.
.
.
.
.
.
(
1
)
\Delta V = \; 1.67 \text{m}^3 \; \; . . . . . .(1)
Δ
V
=
1.67
m
3
......
(
1
)
Δ
Q
=
Δ
U
+
w
\Delta Q = \; \Delta U + w
Δ
Q
=
Δ
U
+
w
m
L
v
=
Δ
U
+
(
1.013
×
1
0
5
)
(
1.67
)
mL_v = \; \Delta U + (1.013 \times 10^5) (1.67)
m
L
v
=
Δ
U
+
(
1.013
×
1
0
5
)
(
1.67
)
Δ
U
=
(
2257
−
170
)
1
0
3
\; \Delta U = (2257-170) 10^3
Δ
U
=
(
2257
−
170
)
1
0
3
Δ
U
=
2090
kJ
(approx.)
Ans. Option
→
4
\; \Delta U = 2090 \text{kJ} \; \text{ (approx.) } \; \; \; \text{ Ans. Option } \; \rightarrow 4
Δ
U
=
2090
kJ
(approx.)
Ans. Option
→
4
© examsnet.com
Go to Question:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
Prev Question
Next Question
More Free Exams
AIEEE Previous Papers
BITSAT Exam Previous Papers
JEE Adv
JEE Advanced Model Papers
JEE Advanced Previous Papers
JEE Main PYQ
JEE Mains Model Papers
VITEEE Previous Papers