Class 12-science H C VERMA Solutions Physics Chapter 6 - Heat Transfer
Heat Transfer Exercise 98
Solution 1
We know,
Solution 2
We know,
= 440W
Solution 3
We know,
Solution 4
We know,
Also Rate of conversion = 
Solution 5
We know,
Solution 6
We know,
Ice melt at rate of m/t which is:
Solution 7
We know,
Also
Solution 8
We know,
Solution 9
We know,
= 6000
Heat goes out from two surfaces, Hence net heat flow is:
= 12000
Also,
Solution 10
(a) We know,
________(1)
Also,
(b) From equation (1) heat flow,
Heat Transfer Exercise 99
Solution 11
Let the point touched be 0. As no heat is flowing
x is distance between 0°C end and point which is touched.
Solution 12
We know,
Solution 13
We know,
Solution 14
We know,
Rearranging
(As Q = mL)
)
(b) Now, we consider a small ship of ice.
And, 
Or 
Or
Integrating
Solution 15
Heat conducted at point above which ice cannot be formed is same from both levels. Thus,
Solution 16
(a) We know
(b) 
(c) 
Solution 17
We know,
And
Taking ratio we get
Solution 18
We know,
And 
Net 
Also, 
Solution 19
We know,
Solution 20
We know,
Now let 
Solution 21
We know,
Cylinders are concentric and heat flow through them is:
Rearranging and integrating
Solution 22
In this case thermal conductivities are in series with each other. Thus,
Solving we get
Solution 23
Both elements are in series with each other. Thus,
Solution 24
Since rod of the both elements are joined in parallel. Thus,
Also,
Heat Transfer Exercise 100
Solution 25
Net heat per second is given as:
+ 
Heat drawn per minute is
Solution 26
According to Question
Also,
And,
Taking ratio, we get
= 
= 
________(2)
Substituting equation (2) in (1) we get
Solution 27
For bent part,
For straight part,
Dividing (1) and (2) we get,
Solution 28
(a) We know,
= 
(b) Resistance by glass is, 
And resistance by air is 
We know,
Solution 29
For case I:
And
Here
For case II:
And
Here
Solution 30
Net Resistance is:
Also,
(b) Now,
= 
= 
Also,
(c) 
Also,
Solution 31
According to Question,
Solution 32
In bar F, temperature at both ends is same:
As
Solution 33
We Know,
And
Simplifying and differentiating w.r.t 'x'
- 0
Also,
Integrating
Heat Transfer Exercise 101
Solution 34
We Know,
Also,
Solving we get
Solution 35
For small strip having thickness 'dr', we have,
Rearranging and integrating we get,
Solving we get
Solution 36
We know,
Rise in temperature is : 
And Fall in temperature is : 
Also, Final temperatures are given as :
And
Now, net temperature is
= 
Rearranging and integrating we get
Solution 37
We know,
Rise in temperature is : 
And Fall in temperature is : 
Final temperature are given as:
And
And also,
Solving we get
Integrating we get
Solution 38
We know,
Also
Integrating and solving we get
And 
Solution 39
Energy radiated is given as:
Solution 40
We know,
Solution 41
(a) Heat loss is the energy radiated
Ratio is 1:4
(b) Emissivity is same in both cases
Ratio is 2.9:1
Solution 42
We know,
Solution 43
(a) We know,
(b) Also,
Solution 44
We know
Solution 45
Let us assume that cube is a black body
Thus,
Solution 46
We know
In case of anybody:
And in case of black body:
Dividing equation (1) and (2) we get
Solution 47
KA is high so it will conduct all heat while KB is low so it will so it will act as poor conductor.
Also 
Similarly 
Solution 48
We know,
Also,
From Equation (1) and (2)
Heat Transfer Exercise 102
Solution 49
We know,
Also,
Solution 50
In case of water,
In case of Kerosene,
Equating (1) and (2) we get
Solution 51
I st CASE:
Average temperature is given as:
And Average temperature that is different from surroundings is:
Now,
Rate of fall of temperature = 
And also Rate of fall of temperature
= 
__________(1)
II nd CASE:
Average temperature is given as:
And Average temperature that is different from surroundings is:
Now,
Rate of fall of temperature = 
And also Rate of fall of temperature
= 
Solution 52
We know,
Rate of heat flow is given as:
And
Solution 53
(a) We know,
Also,
P = H (As in steady state there is no loss of heat)
(b) We know,
(By Newton's law of cooling)
Now again,
(c) Here net heat absorbed is:
Solution 54
(a) Power of heater is given as:
(b) Power Radiated is given as:
(c) We know,
And
Solution 55
(a) Maximum heat loss by body is given as:
(b) If 90% of max. heat is lost then, temperature decrease is:
Temperature at 
Also
Integrating , we get
(From (1))
