Question 397: Properties of Matter

Question

Consider two rods of same length and different specific heats (S
1
, S
2
), conductivities (K
1
, K
2
) and area of cross-sections (A
1
, A
2
) and both having temperatures T
1
and T
2
at their ends. If rate of loss of heat due to conduction is equal, then

Accepted Answer

Rate of heat loss in rod 1 = Q
1
$$ = {{{K_1}{A_1}\left( {{T_1} - {T_2}} ight)} \over {{l_1}}}$$
Rate of heat loss in rod 2 = Q
2
$$ = {{{K_2}{A_2}\left( {{T_1} - {T_2}} ight)} \over {{l_2}}}$$
By problem, Q1 = Q2.
$$ 	herefore {{{K_1}{A_1}\left( {{T_1} - {T_2}} ight)} \over {{l_1}}} = {{{K_2}{A_2}\left( {{T_1} - {T_2}} ight)} \over {{l_2}}}$$
$$ 	herefore $$  K
1
A
1
= K
2
A
2
.   $$\left[ \because{{l_1} = {l_2}} ight]$$

Answer

K
1
A
1
= K
2
A
2

Answer

$${{{K_1}{A_1}} \over {{S_1}}} = {{{K_2}{A_2}} \over {{S_2}}}$$

Answer

K
2
A
1
= K
1
A
2

Answer

$${{{K_2}{A_1}} \over {{S_2}}} = {{{K_1}{A_2}} \over {{S_1}}}$$

Properties of Matter

Consider two rods of same length and different specific heats (S 1 , S 2 ), conductivities (K 1 , K 2 ) and area of cross-sections (A 1 , A 2 ) and both having temperatures T 1 and T 2 at their ends. If rate of loss of heat due to conduction is equal, then

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