2019-03-01_Physics_Times

3

(8^33 )^7

3 3

C V C

 V V 

3

2

7 7 7

3 3 3

V P PV RT

W

V

 

   

 

The process AB can be expressed by a general

equation.

P a bV  (1)

where a and b are constants

Now PV nRT (2)

From eq’s (1) & (2)

T^1 aV bV^2

nR

   

When T is maximum

0

dT

dV



 

1

2 0

dT

a bV

dV nR

  

V a b / 2

Let C is a point on the line AB where T is maximum.

From A C gas absorbs heat and from C B

it rejects heat.

If Q 1 is heat absorbed and Q 2 heat rejected then

2 2

1 1

1 1

Q T

Q T



   

      

   

Given that PVConstant

nRTV TV (^1) Constant
V
  
T 1V dV V dT^2  ^1  0
 ^1
dV V
dT T 


 (1)
From first law of thermodynamics
dQ dU dW 
nCdT nC dT PdV V 
V
PdV
C C
ndT
  (2)
From eq’s (1) & (2)
V 1
R
C C

 

The possible correct option is (c).
Two bottles are shown in the figure
R R RA 0 , B2 ,R h h h 0 A 0 & B 2 h 0
The rate of heat loss is
dQ
e A T
dt
 
mcdT
e A T
dt
 
dT kA
dt m

m
dt dT
kA

Where k is a constant
As dT is same for both bottles
(^) ..
V
t k k h
A

  
19.Sol:
20.Sol:
Heat Transfer
1.Sol:
2.Sol:
0
0
2
2
A A
B A
B B
t h h
t t
t h h
   
Let T is the temperature of the junction
The net heat inflow at the junction is equal to the
net heat outflow.