Question

In a composite rod, when two rods of different lengths and the same area of cross-section, are joined end to end then if $K$ is the effective coefficient of thermal conductivity $\dfrac{{{l_1} + {l_2}}}{K}$ is equal to
A. $\dfrac{{{l_1}}}{{{K_1}}} - \dfrac{{{l_2}}}{{{K_2}}}$
B. $\dfrac{{{l_1}}}{{{K_2}}} - \dfrac{{{l_2}}}{{{K_1}}}$
C. $\dfrac{{{l_1}}}{{{K_1}}} + \dfrac{{{l_2}}}{{{K_2}}}$
D. $\dfrac{{{l_1}}}{{{K_2}}} + \dfrac{{{l_2}}}{{{K_1}}}$

Answer 1

Initial temperature difference between the two rods $= {T_1} - {T_2} = {T_1} - {T_2} + \left( {T - T} \right)$ . Use the equation for heat transfer is $H = \dfrac{{KA(\Delta T)}}{l}$ for the first rod, i.e. $H = \dfrac{{{K_1}A\left( {{T_1} - T} \right)}}{{{l_1}}}$ , the second rod, ie. $H = \dfrac{{{K_2}A\left( {T - {T_2}} \right)}}{{{l_2}}}$ and the whole combination i.e. $H = \dfrac{{KA\left( {{T_1} - {T_2}} \right)}}{{{l_1} + {l_2}}}$ . Put these values in the equation ${T_1} - {T_2} = \left( {{T_1} - T} \right) + \left( {T - {T_2}} \right)$ to reach the solution.

Complete step by step answer:
Considering at the time of joining the two rods, the rods were at a temperature ${T_1}$ and ${T_2}$ respectively. Also, let the final temperature of both the rods at equilibrium be $T$ .
Now the difference between the initial temperatures of the two rods $= {T_1} - {T_2} = {T_1} - {T_2} + \left( {T - T} \right)$
$$\Rightarrow {T_1} - {T_2} =$$$ \Rightarrow {T_1} - {T_2} = \left( {{T_1} - T} \right) + \left( {T - {T_2}} \right)$(Equation 1) Now we know that the heat will be transferred across the two rods when they are joined together. The equation of heat transfer is$H = \dfrac{{KA(\Delta T)}}{l}$$H = $Heat transfer between the rods$K = $Thermal coefficient of the material$A = $Area of the cross-section of the body$\Delta T = $The temperature difference between the initial and final temperature of the body$l = $Length of the body So for the first body, the equation of heat becomes$H = \dfrac{{{K_1}A\left( {{T_1} - T} \right)}}{{{l_1}}}$$ \Rightarrow {T_1} - T = \dfrac{{{l_1}H}}{{{K_1}A}}$(Equation 2) So for the second body, the equation of heat becomes$H = \dfrac{{{K_2}A\left( {T - {T_2}} \right)}}{{{l_2}}}$$ \Rightarrow T - {T_2} = \dfrac{{H{l_2}}}{{{K_2}A}}$(Equation 3) The equation for the heat transfer in combination is given by$H = \dfrac{{KA\left( {{T_1} - {T_2}} \right)}}{{{l_1} + {l_2}}}$$ \Rightarrow {T_1} - {T_2} = \dfrac{{\left( {{l_1} + {l_2}} \right)H}}{{KA}}$(Equation 4) Put the values from equation 2, equation 3, and equation 4 in equation 1, we get$\dfrac{{\left( {{l_1} + {l_2}} \right)H}}{{KA}} = \dfrac{{{l_1}H}}{{{K_1}A}} + \dfrac{{H{l_2}}}{{{K_2}A}}$$ \Rightarrow \dfrac{{\left( {{l_1} + {l_2}} \right)}}{K} = \dfrac{{{l_1}}}{{{K_1}}} + \dfrac{{{l_2}}}{{{K_2}}}$

So, the correct answer is “Option C”.

Note:
A material's thermal conductivity is a measure of its ability to conduct heat. Heat transfer occurs in low thermal conductivity materials at a lower rate than in high thermal conductivity materials. Metals, for example, usually have high thermal conductivity and are very efficient at heat conduction, whereas insulating materials such as Styrofoam are the opposite.