Question 349: Thermodynamics & KTG

Question

Three rods of identical cross-section and lengths are made of three different materials of thermal
conductivity K
1
, K
2
and K
3
, respecrtively. They are joined together at their ends to make a long rod
(see figure). One end of the long rod is maintained at 100
o
C and the other at 0
o
C (see figure). If
the joints of the rod are at 70
o
C and 20
o
C in steady state and there is no loss of energy from the
surface of the rod, the correct relationship between K
1
, K
2
and K
3
is :

Accepted Answer

All three rods have same length($$l$$) and area of cross-section(A).
They are in series combination so heat current is same for all rods.
$$ 	herefore $$ $${\left( {{{\Delta Q} \over {\Delta t}}} ight)_1} = {\left( {{{\Delta Q} \over {\Delta t}}} ight)_2} = {\left( {{{\Delta Q} \over {\Delta t}}} ight)_3}$$
$$ \Rightarrow $$ $${{\left( {100 - 70} ight){K_1}A} \over l} = {{\left( {70 - 20} ight){K_2}A} \over l} = {{\left( {20 - 0} ight){K_3}A} \over l}$$
$$ \Rightarrow $$ 30K
1
= 50K
2
= 20K
3
$$ 	herefore $$ K
1
: K
3
= 2 : 3,
K
2
: K
3
= 2 : 5

Answer

K
1
: K
3
= 2 : 3,
K
2
: K
3
= 2 : 5

Answer

K
1
< K
2
< K
3

Answer

K
1
: K
2
= 5 : 2,
K
1
: K
3
= 3 : 5

Answer

K
1
> K
2
> K
3

Thermodynamics & KTG

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