Examsnet
Unconfined exams practice
Home
Exams
Banking Entrance Exams
CUET Exam Papers
Defence Exams
Engineering Exams
Finance Entrance Exams
GATE Exam Practice
Insurance Exams
International Exams
JEE Exams
LAW Entrance Exams
MBA Entrance Exams
MCA Entrance Exams
Medical Entrance Exams
Other Entrance Exams
Police Exams
Public Service Commission (PSC)
RRB Entrance Exams
SSC Exams
State Govt Exams
Subjectwise Practice
Teacher Exams
SET Exams(State Eligibility Test)
UPSC Entrance Exams
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 Question Papers
NCERT Books
NCERT Exemplar Books
NCERT Study Notes
CBSE Study Concepts
CBSE Class 10 Solutions
CBSE Class 12 Solutions
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
Careers
Contact Us
Our Apps
Privacy
Test Index
JEE Advanced 2020 Paper 2
Show Para
Hide Para
Share question:
© examsnet.com
Question : 18
Total: 54
A container with 1 kg of water in it is kept in sunlight, which causes the water to get warmer than the surroundings. The average energy per unit time per unit area received due to the sunlight is 700
W
m
−
2
and it is absorbed by the waterover an effective area of
0.05
m
2
. Assuming that the heat loss from the water to the surroundings is governed by Newton's law of cooling, the difference (in
‌
°
C
) in the temperature of water and the surroundings aftera long time will be _____. (Ignore effect of the container, and take constant for Newton's law of cooling
=
0.001
s
−
1
,
Heat capacity of water =
4200
J
k
g
−
1
K
−
1
)
[JEE Adv 2020 P2]
Your Answer:
Validate
Solution:
In equilibrium
Heat received per sec = Heat loss per sec
E
×
A
=
K
×
m
s
[
θ
−
θ
0
]
×
M
⇒
(
θ
−
θ
0
)
=
E
×
A
K
×
S
×
M
=
700
×
0.05
10
−
3
×
4200
×
1
°
C
=
700
×
5
×
10
−
2
10
−
3
×
42
×
10
2
=
35
4.2
=
350
42
=
50
6
=
8.33
°
C
© 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
Prev Question
Next Question
More Free Exams:
AIEEE Previous Papers
BITSAT Exam Previous Papers
JEE Advanced Chapters Wise Questions
JEE Advanced Model Papers
JEE Mains Chapters Wise Previous Papers
JEE Mains Model Papers
JEE Mains Previous Papers
VITEEE Previous Papers