# The end Q and R of two thin wires, PQ and RS, are soldered: Heat Conduction

Problem: The end Q and R of two thin wires, PQ and RS, are soldered (joined) together. Initially each of the wires has a length of 1 m at 10°C. Now the end P is maintained at 10°C, while the end S is heated and maintained at 400°C. The system is thermally insulated from its surroundings. If the thermal conductivity of wire PQ is twice that of the wire RS and the coefficient of linear thermal expansion of PQ is 1.2×10-5 K-1, the change in length of the wire PQ is  (IIT JEE 2016)

1. 0.78 mm
2. 0.90 mm
3. 1.56 mm
4. 2.34 mm

Solution: Let steady state temperature of the junction (Q,R) be T. In steady state, the rate of heat flow from the end S to R is equal to the rate of heat flow from the end Q to P because the system is insulated from the surrounding. Thus, \begin{align} \frac{\mathrm{d}Q_\text{SR}}{\mathrm{d}t}&=\frac{\mathrm{d}Q_\text{QP}}{\mathrm{d}t}, \quad \text{i.e.,} \nonumber \\ \frac{\kappa A (400-T)}{\Delta x}&=\frac{2\kappa A (T-10)}{\Delta x},\nonumber \end{align} which gives T = 140°C. Let P be the origin and x-axis points towards Q. In steady state, the temperature on the wire varies linearly with the distance x i.e., T(x)=mx+c, where m is the slope and c is the intercept on T axis. In wire PQ, the temperature is T = 10°C at x = 0 m and the temperature is T = 140°C at x = 1 m. Substitute these values in T(x) = mx+c to get c = 10°C and m = 130°C/m. Thus, steady state temperature on the wire PQ is given by \begin{align} T(x)=130x+10. \nonumber \end{align} Now, consider a small element dx of the wire PQ located at a distance x from the end P. Initial temperature of this element was 10°C and final temperature (in steady state) is T(x)=130x+10. The coefficient of linear thermal expansion of PQ is 1.2×10-5K-1. The increase in length of this element due to thermal expansion is given by \begin{align} \mathrm{d}l & =\alpha\Delta T \mathrm{d}x \\ &=\alpha(130x+10-10)\mathrm{d}x \\ &=130\alpha x \mathrm{d}x. \nonumber \end{align} Integrate from x = 0 m to x = 1 m to get increase in length as \begin{align} l&=\int_{0}^{1} 130\alpha x\mathrm{d}x=130\alpha \left[\frac{x^2}{2}\right]_{0}^{1} \nonumber\\ &=130 (1.2\times{10}^{-5})\frac{1}{2}\\ &=0.78\,\mathrm{mm}. \nonumber \end{align} Detailed analysis of heat transfer reveals that temperature obeys the differential equation $\frac{1}{c^2}\frac{\partial T}{\partial t}=\frac{\mathrm{d}^2T}{\mathrm{d}x^2}$, where $c$ is a constant. In steady state, temperature is independent of time i.e., $\frac{\partial T}{\partial t}=0$ which gives $\frac{\mathrm{d}^2T}{\mathrm{d}x^2}=0$. Integrate it twice to get $T(x)=mx+c$.