Two rods, one made of copper and the other made of steel, of the same length and same cross-sectional area are

Question

Two rods, one made of copper and the other made of steel, of the same length and same cross-sectional area are joined together. The thermal conductivity of copper and steel are $385	ext{ J s}^{-1}	ext{K}^{-1}	ext{m}^{-1}$ and $50	ext{ J s}^{-1}	ext{K}^{-1}	ext{m}^{-1}$ respectively. The free ends of copper and steel are held at $100^{\circ}	ext{C}$ and $0^{\circ}	ext{C}$ respectively. The temperature at the junction is, nearly:

Accepted Answer

Let the temperature of the junction be $T$. In the steady state, the rate of heat flow through the copper rod is equal to the rate of heat flow through the steel rod.
$\frac{dQ}{dt} = \frac{K_c A (100 - T)}{L} = \frac{K_s A (T - 0)}{L}$
Since the length ($L$) and cross-sectional area ($A$) of both rods are the same, they cancel out:
$K_c(100 - T) = K_s(T - 0)$
Given $K_c = 385	ext{ J s}^{-1}	ext{K}^{-1}	ext{m}^{-1}$ and $K_s = 50	ext{ J s}^{-1}	ext{K}^{-1}	ext{m}^{-1}$:
$385(100 - T) = 50T$
$38500 - 385T = 50T$
$435T = 38500$
$T = \frac{38500}{435} \approx 88.5^{\circ}	ext{C}$

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