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QJEEAdv09Phde10: JEE Advanced 2009 Physics - Class 11 JEEAdv09Phde10 Solution: A metal rod AB of length 10x has its one end A in ice at 0 C…

A metal rod AB of length 10x has its one end A in ice at 0 ∘^\circ C and the other end B in water at 100 ∘^\circ C. If a point P on the rod is maintained at 400 ∘^\circ C, then it is found that equal amounts of water and ice evaporate and melt per unit time. The latent heat of evaporation of water is 540 cal/g and latent heat of melting of ice is 80 cal/g. If the point P is at a distance of λx\lambda x from the ice end A, find the value of λ\lambda . (Neglect any heat loss to the surrounding.)

Step-by-Step Verified Solution

The steady rate of flow of heat is

ΔQΔt=kA(ΔTΔx){{\Delta Q} \over {\Delta t}} = kA\left( {{{\Delta T} \over {\Delta x}}} \right)

IIT-JEE 2009 Paper 2 Offline Physics - Heat and Thermodynamics Question 14 English Explanation

Therefore, kA(400−0∘)λx=mice lice{{kA(400 - 0^\circ )} \over {\lambda x}} = {m_{ice}}\,{l_{ice}}

kA(400−100)(10−λ)x=mwater lwater{{kA(400 - 100)} \over {(10 - \lambda )x}} = {m_{water}}\,{l_{water}}

where mice{m_{ice}} is the mass of ice melted per unit times and mwater{m_{water}} is the mass of water evaporated per unit times. It is given that

mice=mwater{m_{ice}} = {m_{water}}

Therefore, kA×400λx(lice)=kA×300(10−λ)x lwater{{kA \times 400} \over {\lambda x({l_{ice}})}} = {{kA \times 300} \over {(10 - \lambda )x\,{l_{water}}}}

⇒4(10−λ)×540=3λ×80\Rightarrow 4(10 - \lambda ) \times 540 = 3\lambda \times 80

⇒5400−540λ=60λ\Rightarrow 5400 - 540\lambda = 60\lambda

⇒λ=9\Rightarrow \lambda = 9

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