Question
Asked Dec 25, 2019
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A bar of gold (Au) is in thermal contact with a bar of silver (Ag) of the same length and area (Fig. P11.63). One end of the compound bar is maintained at 80.0°C, and the opposite end is at 30.0°C. Find the temperature at the junction when the energy flow reaches a steady state.

80.0°C
Au
Insulation
Ag
30.0°C
Figure P11.63
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80.0°C Au Insulation Ag 30.0°C Figure P11.63

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Expert Answer

Step 1

Given:

Length and area of gold and silver bars are equal. Temperatures of their end are given. 

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80.0°C Au Insulation Ag 30.0°C T = 80° & T = 30° A =A Au Ag Au L =LAg K = 318 W / mK, Au KA, = 429 W/ mK Ag

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Step 2

Calculating the junction temperature when the energy flow is steady at the junction:

Let the steady state temperature of th...

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KAA (T – T;unction)_ KAA (Trancion -Tag) Junctic Au L Ag Au Substituting: T = 80° & T = 30° A A = A, & LAu =L K =318 W/m K, & K Au Ag = 429 W / mK Au Ag (T, etion – 30°) LAE (429W/ mK)A (318W / mK)A (80° –Trcien) As length and area of the both the rods are equal, (318 W/ mK)(80° -Tmsisn) = (429W /mK)(Tareien -30°) Junctic (747W/ mK)(Tncion) = 38310W/m 38310W/ m 51.28° (Tjurzion) 747W/ mK (Tjugztion) = 51.28° C

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Science

Physics

Heat Transfer