Question 42: Thermodynamics & KTG

Question

A block of metal A is connected in series with another block of metal B such that the two metal blocks have the same area of cross sections. The thermal conductivity of metal A is K and the free end of metal A is at $80^{\circ} \mathrm{C}$. The temperature of the interface is $60^{\circ} \mathrm{C}$ and the free end of metal B is at $20^{\circ} \mathrm{C}$. Assuming the two metals have the same thickness, the conductivity of metal B is:

Accepted Answer

Let the crossÃ¢ÂÂsectional area be $A$ and each block have thickness $L$. In steady state the heat current $I$ is the same through both:
Through metal A:
$    I \;=\; K\,A\;\frac{\Delta T_A}{L}        \;=\; K\,A\;\frac{80-60}{L}        \;=\; K\,A\;\frac{20}{L}.    $
Through metal B (with conductivity $k_B$):
$    I \;=\; k_B\,A\;\frac{\Delta T_B}{L}        \;=\; k_B\,A\;\frac{60-20}{L}        \;=\; k_B\,A\;\frac{40}{L}.    $
Setting the two heat currents equal,
$ K\,A\,\frac{20}{L} \;=\; k_B\,A\,\frac{40}{L} \quad\Longrightarrow\quad k_B \;=\;\frac{20}{40}\,K \;=\;\frac{K}{2}. $
Answer: $\displaystyle \frac{K}{2}$  (Option C)

Answer

$2 K$

Answer

$4 K$

Answer

$\frac{K}{2}$

Answer

$\frac{K}{4}$

Thermodynamics & KTG

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