Question

Three copper rods and three steel rods each of length $l$ = 10 cm and area of cross - section $1c{m}^2$ are connected as shown If ends A and E are maintained at temperatures ${125}^{\circ }$C and $o^{\circ }$C respectively, calculate the amount of heat flowing per second from the hot to cold function. $[K_{cu}=400W/m-K,K_{steel}=50W/m-K]$

• A. 2 watt
• B. 9 watt
• C. 4 watt
• D. 10 watt

Answer 1

The correct option is C 4 watt

$R_{steel}=\frac{L}{KA}=\frac{{10}^{-1}m}{50\left(w/m{-}^{\circ }C\right)\times {10}^{-4}{m}^2}=\frac{1000}{50}{}^{\circ }C/W$
Similarly $R_{c}u=\frac{1000}{50}{}^{\circ }C/W$
Junction C and 0 are identical in every respect and both will have same temperature. Consequently.the rod CD is in thermal equilibrium and no heat will flow through it. Hence it can be neglected in further analysis.Now rod BC and CE are in series their equivalent resistance is $R_1=R_s+R_{cu}$ similarly rods SO and DE are in series with same equivalent resistance $R_1=R_s+R_{cu}$ these two are in parallel giving an equivalent resistance of
$\frac{R_1}{2}=\frac{R_s+R_{cu}}{2}$
This resistance is connected in series with rod AB. Hence the net equivalent of the combination is
$R=R_{steel}+\frac{R_1}{2}=\frac{3R_{steel}+R_{cu}}{2}=500{(\frac{3}{50}+\frac{1}{400})}^{\circ }C/W$
Now $i=\frac{T_H-T_c}{R}=\frac{{125}^{\circ }C}{500(\frac{3}{50}+\frac{1}{400}){}^{\circ }C/W}=4watt$