Question

Three cubes of equal side length of steel, lead, and copper, are arranged between heat reservoirs at 150C and 0C as shown in the diagram. The heat current between the reservoirs is 175 W.
a) What is the length of the side of a cube?
b) What is the temperature at the junction between the steel and lead cubes?

Answer 1

My thoughts so far:
\[H = kA\frac{\Delta T}{L}\]

Wherein the change in temperature is between the two hot rods.

Since there's three cubes, the total length of the system is 3L. The cross-sectional area, A, is L^2.

Answer 2

\[H = kL^2\frac{\Delta T}{3L}\]
\[H = kL\frac{\Delta T}{3}\]

Answer 3

here, we have three temperature changes:
\[\Delta {T_1} = \frac{{WL}}{{k_{steel} A}} = \frac{W}{{k_{steel} L}}\]
being cross sectional area A such that \(A=L^2\)
then we can write the remaining temperature changes as below:
\[\begin{gathered}
\Delta {T_3} = \frac{W}{{{k_{copper}}L}} \hfill \\
\hfill \\
\Delta {T_2} = \frac{W}{{{k_{lead}}L}} \hfill \\
\end{gathered} \]

Answer 4

then, if I call with \(\Delta T_0=150-0=150\), we can write this:
\[\large \Delta {T_0} = \Delta {T_1} + \Delta {T_2} + \Delta {T_3} = \frac{W}{L}\left( {\frac{1}{{{k_{steel}}}} + \frac{1}{{{k_{lead}}}} + \frac{1}{{{k_{copper}}}}} \right)\]

Answer 5

\[\Delta {T_0} = \Delta {T_1} + \Delta {T_2} + \Delta {T_3} = \frac{W}{L}\left( {\frac{1}{{{k_{steel}}}} + \frac{1}{{{k_{lead}}}} + \frac{1}{{{k_{copper}}}}} \right)\]

Answer 6

where \(W\) is the heat current

Answer 7

Thank you again for your help!