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PhysicsInteger+4 / -1Question 50 of 57

IIT-JEE 2009 · Paper 2, Physics question 50

A metal rod AB of length 10x10x has its one end A in ice at 0°C and the other end B in water at 100°C. If a point P on the rod is maintained at 400°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.]

Answer and worked solution

Official answer: 9

Heat currents from P (400°C): to the ice end over λx\lambda x, HA∝400λxH_A\propto\dfrac{400}{\lambda x}; to the water end over (10−λ)x(10-\lambda)x, HB∝300(10−λ)xH_B\propto\dfrac{300}{(10-\lambda)x}.

Equal masses per unit time. HA80=HB540\dfrac{H_A}{80}=\dfrac{H_B}{540}:
40080λ=300540(10−λ)  ⟹  5λ=59(10−λ)  ⟹  λ=9.\frac{400}{80\lambda}=\frac{300}{540(10-\lambda)}\implies\frac5\lambda=\frac5{9(10-\lambda)}\implies\lambda=9.

Answer: 9

Topics:ConductionLatent heat and phase change
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