Transmission Of Heat Question 103
Question: A composite metal bar of uniform section is made up of length 25 cm of copper, 10 cm of nickel and 15 cm of aluminium. Each part being in perfect thermal contact with the adjoining part. The copper end of the composite rod is maintained at $ 100^{o}C $ and the aluminium end at $ 0^{o}C $ . The whole rod is covered with belt so that there is no heat loss occurs at the sides. If $ {K _{\text{Cu}}}=2K _{Al} $ and $ K _{Al}=3{K _{\text{Ni}}} $ , then what will be the temperatures of $ Cu-Ni $ and $ Ni-Al $ junctions respectively
Options:
A) $ {{23.33}^{o}}C $ and $ A $
B) $ {{83.33}^{o}}C $ and $ 20^{o}C $
C) $ 50^{o}C $ and $ 30^{o}C $
D) $ 30^{o}C $ and $ 50^{o}C $
Show Answer
Answer:
Correct Answer: B
Solution:
If suppose $ K _{Ni}=K\Rightarrow K _{Al}=3K $ and $ K _{Cu}=6K. $ Since all metal bars are connected in series So $ {{( \frac{Q}{t} )} _{Combination}}={{( \frac{Q}{t} )} _{Cu}}={{( \frac{Q}{t} )} _{Al}}={{( \frac{Q}{t} )} _{Ni}} $ and $ \frac{3}{K _{eq}}=\frac{1}{K _{Cu}}+\frac{1}{K _{Al}}+\frac{1}{K _{Ni}}=\frac{1}{6K}+\frac{1}{3K}+\frac{1}{K}=\frac{9}{6K} $
Therefore $ K _{eq}=2K $ Hence, if $ {{( \frac{Q}{t} )} _{Combination}}={{( \frac{Q}{t} )} _{Cu}} $
Therefore $ \frac{K _{eq},A(100-0)}{l _{Combination}}=\frac{K _{Cu}A(100-{\theta _{1}})}{l _{Cu}} $
Therefore $ \frac{2K,A,(100-0)}{(25+10+15)}=\frac{6K,A,(100-{\theta _{1}})}{25} $
Therefore $ {\theta _{1}}=83.33{}^\circ C $ Similar if $ {{( \frac{Q}{t} )} _{Combination}}={{( \frac{Q}{t} )} _{Al}} $
Therefore $ \frac{2K,A(100-0)}{50}=\frac{3K,A({\theta _{2}}-0)}{15} $
Therefore $ {\theta _{2}}=20^{o}C $
BETA
