Question #161723

1.  At 00C, a rod of aluminum and a rod of copper have the same length, 1.25 meters. To what common temperature must they be heated to differ in length by 0.0750 millimeter?



1
Expert's answer
2021-02-07T19:22:37-0500

By the definition of thermal linear expansion, we have:


LAlLAl,0LAl,0=αAl(tcomm0C),\dfrac{L_{Al}-L_{Al,0}}{L_{Al,0}}=\alpha_{Al}(t_{comm}-0^{\circ}C),LCuLCu,0LCu,0=αCu(tcomm0C).\dfrac{L_{Cu}-L_{Cu,0}}{L_{Cu,0}}=\alpha_{Cu}(t_{comm}-0^{\circ}C).

Then, taking into account that LAl,0=LCu,0=L0L_{Al,0}=L_{Cu,0}=L_0 we can rewrite our equations as follows:


LAlL0=αAlL0tcomm,L_{Al}-L_0=\alpha_{Al}L_0t_{comm},LCuL0=αCuL0tcomm.L_{Cu}-L_0=\alpha_{Cu}L_0t_{comm}.

Subtracting second equation from the first one, we get:


LAlLCu=(αAlαCu)L0tcomm,L_{Al}-L_{Cu}=(\alpha_{Al}-\alpha_{Cu})L_0t_{comm},tcomm=LAlLCu(αAlαCu)L0,t_{comm}=\dfrac{L_{Al}-L_{Cu}}{(\alpha_{Al}-\alpha_{Cu})L_0},tcomm=7.5105 m(24106 1C17106 1C)1.25 m=8.57C.t_{comm}=\dfrac{7.5\cdot10^{-5}\ m}{(24\cdot10^{-6}\ \dfrac{1}{^{\circ}C}-17\cdot10^{-6}\ \dfrac{1}{^{\circ}C})\cdot1.25\ m}=8.57^{\circ}C.

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Comments

Blessed Joy
08.02.21, 14:44

Noted. Thank you very much!! God bless

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