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The Three Rods Shown in Figure (28−E7) Have Identical Geometrical Dimensions. Heat Flows from the Hot End at a Rate of 40 W in the Arrangement (A). - Physics

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Question

The three rods shown in figure  have identical geometrical dimensions. Heat flows from the hot end at a rate of 40 W in the arrangement (a). Find the rates of heat flow when the rods are joined as in arrangement (b) and in (c). Thermal condcutivities of aluminium and copper are 200 W m−1°C−1 and 400 W m−1°C−1 respectively.

Solution

For arrangement (a),

Temperature of the hot end ,T1 = 100°C

Temperature of the cold end ,T2 = 0°C

`R_s = R_1 + R_2 + R_3`

`(l)/(K_{AI}A)  + l/(K_{cu}A) + {l}/(K_{Al}A)`

`(l)/(A) ( 1/200 +1/400 + 1/200)`

`1/A (5/400)`

`1/Axx 1/80`

`(dQ)/d = q = Rate  of  flow  of  heat = (T_1 - T_2)/R_s `

`= (100 - 0)/((1/Axx1/(80))`             

Given :

q = 40W

`40 = (100)/(1/Axx 1/80)`

`⇒ l/A = 200`

`⇒ l/A = 1/200 `

For arrangement (b),

`"R_net" = R_[AI} + (R_cuxxR_{AI})/(R_cu = R_{AI)}`

      
`= l / (K_AI A) + (l/(K_cuA)xxl/K_(AI))/(l / (K_{cu}A9 )+ (l)/ (K_{AL}A)`

`= (l)/(A.K_Al) +( l )/ (A(K_AL + K_cu)) `

`= l/A  (1/200 + 1/"200 + 400")`

`=l/A ( 1/200 + 1/600)`

= `4/600 1/A`

Rate of flow of heat is given by 

`q = (T_1 - T_2)/ R_"net"`

 `= ((100 - 0))/ (4 l)   600 A`

`= (100xx600)/4 xx1/200`

= 75 W

For arrangement (c),


 `1 / (R"net")  = (1)/(R_{AI)) + 1/R_(cu)+ 1/(R_AI) `

`= K_{K_{AI}}/ (l)+ K_(cuA)/l  + (K_{Al}A)/l`

`1/R_"net" = (K_Al + K_( cu) + K_(Al))A/l`

`l /R_"(net)" = (200 + 400 +200)`

`1/ R_"net" = 200/800`

`=(100)/(1/4)`

rate of heat flow = `(DeltaT)/R_"net"`

`= 100/1/4`400 W

  Is there an error in this question or solution?

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Solution The Three Rods Shown in Figure (28−E7) Have Identical Geometrical Dimensions. Heat Flows from the Hot End at a Rate of 40 W in the Arrangement (A). Concept: Thermal Expansion of Solids.
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