Question 375: Properties of Matter

Question

Consider two rods of same length and different specific heats (S1, S2), conductivities (K1, K2) and area of cross-sections (A1, A2) and both having temperatures T1 and T2 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

K1A1 = K2A2

Answer

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

Answer

K2A1 = K1A2

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 (S1, S2), conductivities (K1, K2) and area of cross-sections (A1, A2) and both having temperatures T1 and T2 at their ends. If rate of loss of heat due to conduction is equal, then

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