Consider two cylindrical metal rods, one lead the other aluminum, connected in series. The temperature at the?

Consider two cylindrical metal rods, one lead the other aluminum, connected in series. The temperature at the lead end of the rods is 20.0 °C; the temperature at the aluminum end is 72.4 °C. Given that the temperature at the lead-aluminum interface is 50.0 °C, and that the lead rod is 14 cm long, find the length of the aluminum rod. cm


Hint: the rate of flow of heat through both rods is the same.





I dont understand physics too well so please try to provide details if you could with ans answer! Thanks!

Consider two cylindrical metal rods, one lead the other aluminum, connected in series. The temperature at the?
..........L...E....A...D%26gt;%26lt;A..L.U.M..N%26gt;


%26lt;%26lt;Q O========OO-------------O %26lt;%26lt;Q


.. 20C^........... .50C^...... 72.5 C^


........%26lt;--- 0.14m ---%26gt;%26lt;----- La ----%26gt;





rods are connected in series, rate of heat flow (like current in series) will be same





Q = conducted heat /sec = - k A ∆T/∆x


--------------------------------------...


k = thermal conductivity, A=cross-section area=pi r^2


∆T = fall in temp = - (T-hot - T-cold)


∆x = L


-------------------------------------


(Q)alm = (Q)lead


ka Aa (∆T)a / La = kl Al (∆T)l / Ll


--------------------------------------...


Aa = Al =same diameter


(∆T)a = (72.5-50) = 22.5 C, (∆T)l = (50-20) = 30 C


Ll = 0.14 : from NET :


ka = 237 W/m-C


kl = 35.3 (do)


-------------------------------------


ka (∆T)a / La = kl (∆T)l / Ll


237*22.5/La = 35.3*30/0.14


La = 237*22.5*0.14 / 35.3*30 = 0.70495 meter


La = length of aluminium rod = 70.5 cm


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