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
Certifications
Technical
Cloud Tech Certifications
Security Tech Certifications
Management
IT Infrastructure
More
About
Careers
Contact Us
Our Apps
Privacy
Test Index
KVPY SA Exam 05-Nov-2017 Question Paper
Show Para
Hide Para
Share question:
© examsnet.com
Question : 23
Total: 80
Ice in a freezer is at –7°C. 100 g of this ice is mixed with 200 g of water at 15°C. Take the freezing temperature of water to be 0°C, the specific heat of ice equal to 2.2 J/g °C, specific heat of water equal to 4.2 J/g °C, and the latent heat of ice equal to 335 J/g. Assuming no loss of heat to the environment, the mass of ice in the final mixture is closest to
88 g
67 g
54 g
45 g
Validate
Solution:
Let mgm of ice melted
Mixture will be at 0ºC
Heat given = 200 × (4.2) (15) J
Heat absorbed = 100 × 2.2 × (7) + m(335)
by solving m = 33 gm
So ice remain = 100 – 33 = 67g
© 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
Prev Question
Next Question