Question

Consider a pair of insulating blocks with thermal resistances $R_1$ and $R_2$ as shown in the figure. The temperature $\theta$ at the boundary between the two blocks is

• A. $\left(\theta_{1}\,\theta_{2}v R_1 R_2\right)/$ $\left(\theta_{1}+\theta_{2}\right)\left(R_{1}+R_{2}\right)$
• B. $\left(\theta_{1}R_{1}+\theta_{2}R_{2}\right)/\left(R_{1}+R_{2}\right)$
• C. $\left[\left(\theta_{1}+\theta_{2}\right)R_{1}R_{2}\right]/\left(R^{2}_{1}+R^{2}_{2}\right)$
• D. $\left(\theta_{1}R_{2}+\theta_{2}R_{1}\right)/\left(R_{1}+R_{2}\right)$

Answer 1

Answer: (D)

$\left(\theta_{1}R_{2}+\theta_{2}R_{1}\right)/\left(R_{1}+R_{2}\right)$

Answer 2

Rate of transmission of heat
$=\frac{Temperature\,difference}{Thermal\,Resistance}$
$\therefore \frac{dQ}{dt}=\frac{d\theta}{R}$
Here, $\frac{dQ}{dt}=\frac{\left(\theta-\theta_{2}\right)}{R_{2}}=\frac{\theta_{1}-\theta }{R_{1}}$
$\Rightarrow \frac{\theta -\theta_{2}}{R_{2}}=\frac{\theta _{1}-\theta }{R_{1}}$
$\Rightarrow R_{1}\theta-R_{1}\theta_{2}=R_{2}\theta_{1}-R_{2}\theta$
$\Rightarrow \theta\left(R_{1}+R_{2}\right)=R_{2}\theta_{1}+R_{1}\theta_{2}$
$\therefore \theta=\frac{\left(R_{2}\theta_{1}+R_{1}\theta_{2}\right)}{\left(R_{1}+R_{2}\right)}$