Transmission Of Heat Question 325
Question: A rod of length I (laterally thermally insulated) of uniform cross-sectional area A consists of a material whose thermal conductivity varies with temperature as $ K=\frac{k _{0}}{a+bT}, $ where, $ k _{0}, $ a and b are constants. $ T _{1} $ and $ T _{2}(<T _{1}) $ are the temperature of two ends of rod. Then, rate of flow of heat across the rod is
Options:
A) $ \frac{Ak _{0}}{bl}[ \frac{a+bT _{1}}{a+bT _{2}} ] $
B) $ \frac{Ak _{0}}{bl}[ \frac{a+bT _{2}}{a+bT _{1}} ] $
C) $ \frac{Ak _{0}}{bl}\ln [ \frac{a+bT _{1}}{a+bT _{2}} ] $
D) $ \frac{Ak _{0}}{al}\ln [ \frac{a+bT _{2}}{a+bT _{1}} ] $
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Answer:
Correct Answer: C
Solution:
[c] Idea Rate of flow of heat is given by $ \frac{dQ}{dt}=-KA\frac{dT}{dx} $ . As we know that, $ \frac{dQ}{dt}=-KA\frac{dT}{dx} $ $ \frac{dQ}{dt}=-\frac{k _{0}A}{a+bT}\frac{dT}{dx} $ On integrating both sides within the proper limits. $ \frac{dQ}{dt}\int _{0}^{l}{dx}=-k _{0}A\int _{T _{1}}^{T _{2}}{\frac{dT}{dx}} $ This gives $ \frac{dQ}{dt}=\frac{Ak _{0}}{bl}\ln [ \frac{a+bT _{1}}{a+bT _{2}} ] $ TEST Edge Question related to equivalent thermal conductivity of two or more rods in series and parallel at various temperature can be asked. In series equivalent conductivity is given by $ K _{eq}=\frac{K _{1}K _{2}(L _{1}+L _{2})}{(L _{1}K _{1}+L _{2}K _{2})} $ In parallel equivalent conductivity is given by $ K _{eq}=( \frac{K _{1}A _{1}+K _{2}A _{2})}{A _{1}+A _{2}} ) $
BETA
