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Two rods, one made of copper and the other steel of the same length and cross sectional area are joined together. (The thermal conductivity of copper is $385 \mathrm{~J} \cdot \mathrm{s}^{-1} \cdot \mathrm{m}^{-1} \cdot \mathrm{K}^{-1}$ and steel is $50 \mathrm{~J} \cdot \mathrm{s}^{-1} \cdot \mathrm{m}^{-1} \cdot \mathrm{K}^{-1}$.) If the copper end is held at $100^{\circ} \mathrm{C}$ and the steel end is held at $0^{\circ} \mathrm{C}$, what is the junction temperature (assuming no other heat losses)?
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The correct answer is:
$88^{\circ} \mathrm{C}$
$\frac{100-\mathrm{T}}{\mathrm{R}_{1}}=\frac{\mathrm{T}-0}{\mathrm{R}_{2}}$
$\frac{100-\mathrm{T}}{\mathrm{T}}=\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}$
$\mathrm{R}=\frac{\mathrm{L}}{\mathrm{KA}}$
$\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}$
$\frac{100-\mathrm{T}}{\mathrm{T}}=\frac{50}{385}=\frac{10}{77}$
$7700-77 \mathrm{~T}=10 \mathrm{~T}$
$7700=87 \mathrm{~T}$
$\mathrm{~T}=\frac{7700}{87}=88^{\circ} \mathrm{C}$
$\frac{100-\mathrm{T}}{\mathrm{T}}=\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}$
$\mathrm{R}=\frac{\mathrm{L}}{\mathrm{KA}}$
$\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}$
$\frac{100-\mathrm{T}}{\mathrm{T}}=\frac{50}{385}=\frac{10}{77}$
$7700-77 \mathrm{~T}=10 \mathrm{~T}$
$7700=87 \mathrm{~T}$
$\mathrm{~T}=\frac{7700}{87}=88^{\circ} \mathrm{C}$
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