Question

What is the value of ${\theta }_2$, the temperature of junction C?

• A. 58.2${}^o$C
• B. 40.2${}^o$C
• C. 68.2${}^o$C
• D. 74.2${}^o$C

Answer 1

The correct option is A 58.2${}^o$C

The junction law states that through a junction, The heat flowing in is equal to heat going out. Now we can say the same by differentiating the equation with time and doing that for the rate of flow of heat which is kA(change in temp)/l and we proceed as :

In the steady state there is no accumulation of heat at any point. Hence at B the incoming heat current must be equal to the outgoing heat current.
Let 2k be the thermal conductivity of material x and K the thermal conductivity of material y
$\frac{2KA({90}^{\circ }-{\theta }_1)}{L}=\frac{2KA({\theta }_1-{\theta }_2)}{L}+\frac{KA({\theta }_1-{\theta }_2)}{L}$
$2({90}^{\circ }-{\theta }_1)=2({\theta }_1-{\theta }_2)+({\theta }_1-{\theta }_2)$
${180}^{\circ }=5{\theta }_1-3{\theta }_2$ .....(1)
Again applying the same rule to junction C
$\frac{2KA({\theta }_1-{\theta }_2)}{L}+({\theta }_1-{\theta }_2)\frac{KA}{L}=\frac{KA({\theta }_2-{20}^{\circ })}{L}$
$2({\theta }_1-{\theta }_2)+({\theta }_1-{\theta }_2)={\theta }_2-{20}^{\circ }$
$3{\theta }_1-4{\theta }_2=-{20}^{\circ }$ .....(2)
Solving equation (1) and (2)
${\theta }_1={70.9}^{\circ }$ C
${\theta }_2={58.2}^{\circ }$
Substitue the value of ${\theta }_1$ in the equation obtained
$3\times 60-4{\theta }_2=-20$
${\theta }_2={58.2}^{\circ }$ C