Question

A bar of copper of length $75\text{cm}$ and a bar of steel of length $125\text{cm}$ are joined together end to end. Both are of circular cross-section with diameter $2\text{cm}$. The free ends of the copper and the steel bars are maintained at ${100}^{\circ }\text{C}$ and $0^{\circ }\text{C}$ respectively. The curved surfaces of the bars are thermally insulated. What is the temperature of the $\text{copper - steel}$ junction?
$[\text{Take}:K_{\text{Cu}}=386{\text{Js}}^{-1}{\text{m}}^{-1}{}^{\circ }{\text{C}}^{-1}{K}_{\text{steel}}=46{\text{Js}}^{-1}{\text{m}}^{-1}{}^{\circ }{\text{C}}^{-1}]$

• A. ${63}^{\circ }\text{C}$
• B. ${73}^{\circ }\text{C}$
• C. ${83}^{\circ }\text{C}$
• D. ${93}^{\circ }\text{C}$

Answer 1

The correct option is D ${93}^{\circ }\text{C}$


Let $T$ be the temperature of the junction.

Length, $l_1=75\text{cm}=0.75\text{m}$ and $l_2=125\text{cm}=1.25\text{m}$

In steady state, rate of heat flow will be same in both copper and steel to maintain the constant temperature (with time).

The rate of heat flow is given by,

$\frac{\Delta Q}{\Delta t}=\frac{KA\Delta T}{l}$

$\Rightarrow \frac{K_{\text{Cu}}A(100-T)}{0.75}=\frac{K_{\text{steel}}A(T-0)}{1.25}$

$\Rightarrow \frac{386(100-T)}{0.75}=\frac{46T}{1.25}$

$\Rightarrow 48250-482.5T=34.5T$

$\Rightarrow T\simeq {93}^{\circ }\text{C}$

Hence, option $\left(D\right)$ is correct.

$\text{Alternate solution:}$

Temperature at the junction,

$T=\frac{\frac{K_{cu}{T}_{\text{cu end}}}{l_{cu}}+\frac{K_{steel}{T}_{\text{steel end}}}{l_{steel}}}{\frac{K_{cu}}{l_{cu}}+\frac{K_{steel}}{l_{steel}}}$

$=\frac{\frac{286\times 100}{75}+\frac{46\times 0}{125}}{\frac{286}{75}+\frac{46}{125}}\approx {93.11}^{\circ }\text{C}$

Why this question? Concept: In steady state, heat incoming is equal to heat outgoing.