mortonsalt
  • mortonsalt
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?
Physics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
mortonsalt
  • mortonsalt
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.
mortonsalt
  • mortonsalt
\[H = kL^2\frac{\Delta T}{3L}\] \[H = kL\frac{\Delta T}{3}\]
mortonsalt
  • mortonsalt
1 Attachment

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Michele_Laino
  • Michele_Laino
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: \[\begin{gathered} \Delta {T_3} = \frac{W}{{{k_{copper}}L}} \hfill \\ \hfill \\ \Delta {T_2} = \frac{W}{{{k_{lead}}L}} \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
then, if I call with \(\Delta T_0=150-0=150\), we can write this: \[\large \Delta {T_0} = \Delta {T_1} + \Delta {T_2} + \Delta {T_3} = \frac{W}{L}\left( {\frac{1}{{{k_{steel}}}} + \frac{1}{{{k_{lead}}}} + \frac{1}{{{k_{copper}}}}} \right)\]
Michele_Laino
  • Michele_Laino
\[\Delta {T_0} = \Delta {T_1} + \Delta {T_2} + \Delta {T_3} = \frac{W}{L}\left( {\frac{1}{{{k_{steel}}}} + \frac{1}{{{k_{lead}}}} + \frac{1}{{{k_{copper}}}}} \right)\]
Michele_Laino
  • Michele_Laino
where \(W\) is the heat current
mortonsalt
  • mortonsalt
Thank you again for your help!

Looking for something else?

Not the answer you are looking for? Search for more explanations.