A community for students.
Here's the question you clicked on:
 0 viewing
mortonsalt
 one year ago
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?
mortonsalt
 one year ago
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?

This Question is Closed

mortonsalt
 one year ago
Best ResponseYou've already chosen the best response.0My 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 crosssectional area, A, is L^2.

mortonsalt
 one year ago
Best ResponseYou've already chosen the best response.0\[H = kL^2\frac{\Delta T}{3L}\] \[H = kL\frac{\Delta T}{3}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1here, 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
 one year ago
Best ResponseYou've already chosen the best response.1then, if I call with \(\Delta T_0=1500=150\), 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
 one year ago
Best ResponseYou've already chosen the best response.1\[\Delta {T_0} = \Delta {T_1} + \Delta {T_2} + \Delta {T_3} = \frac{W}{L}\left( {\frac{1}{{{k_{steel}}}} + \frac{1}{{{k_{lead}}}} + \frac{1}{{{k_{copper}}}}} \right)\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1where \(W\) is the heat current

mortonsalt
 one year ago
Best ResponseYou've already chosen the best response.0Thank you again for your help!
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.