JEE Challenger
More from Thermal Properties of Matter
PhyMedAdv 2009 • Paper 2

Distance of Heating Point on Metal Rod for Equal Phase Change

A metal rod AB of length 10x10x has its one end A in ice at 0∘C0^\circ\text{C} and the other end B in water at 100∘C100^\circ\text{C}. If a point P on the rod is maintained at 400∘C400^\circ\text{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/g540\text{ cal/g} and latent heat of melting of ice is 80 cal/g80\text{ 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.]

Official Numerical Answer9

Topics & Concepts

Thermal Properties of MatterHeat TransferCalorimetry

Step-by-Step Solution

To find the value of λ\lambda, we analyze the steady-state heat conduction along the two segments of the metal rod: segment AP\text{AP} and segment PB\text{PB}.

Let:

  • KK be the thermal conductivity of the rod,
  • AA be the uniform cross-sectional area of the rod,
  • TP=400∘CT_{\text{P}} = 400^\circ\text{C} be the temperature of point P\text{P},
  • TA=0∘CT_{\text{A}} = 0^\circ\text{C} be the temperature of end A\text{A} (in ice),
  • TB=100∘CT_{\text{B}} = 100^\circ\text{C} be the temperature of end B\text{B} (in boiling water).

The length of segment AP\text{AP} is:
lAP=λxl_{\text{AP}} = \lambda x

The length of segment PB\text{PB} is:
lPB=(10−λ)xl_{\text{PB}} = (10 - \lambda)x

The rate of heat conduction from point P\text{P} to end A\text{A} is given by Fourier's law of heat conduction:
dQ1dt=KA(TP−TA)lAP=KA(400−0)λx=400KAλx\frac{dQ_1}{dt} = \frac{KA(T_{\text{P}} - T_{\text{A}})}{l_{\text{AP}}} = \frac{KA(400 - 0)}{\lambda x} = \frac{400KA}{\lambda x}

This heat is used to melt ice at 0∘C0^\circ\text{C}. Thus:
dQ1dt=(dmicedt)Lf\frac{dQ_1}{dt} = \left(\frac{dm_{\text{ice}}}{dt}\right) L_f
where Lf=80 cal/gL_f = 80\text{ cal/g} is the latent heat of fusion of ice.

Similarly, the rate of heat conduction from point P\text{P} to end B\text{B} is:
dQ2dt=KA(TP−TB)lPB=KA(400−100)(10−λ)x=300KA(10−λ)x\frac{dQ_2}{dt} = \frac{KA(T_{\text{P}} - T_{\text{B}})}{l_{\text{PB}}} = \frac{KA(400 - 100)}{(10 - \lambda)x} = \frac{300KA}{(10 - \lambda)x}

This heat is used to vaporize water at 100∘C100^\circ\text{C}. Thus:
dQ2dt=(dmwaterdt)Lv\frac{dQ_2}{dt} = \left(\frac{dm_{\text{water}}}{dt}\right) L_v
where Lv=540 cal/gL_v = 540\text{ cal/g} is the latent heat of vaporization of water.

It is given that equal amounts of ice and water melt and evaporate per unit time:
dmicedt=dmwaterdt\frac{dm_{\text{ice}}}{dt} = \frac{dm_{\text{water}}}{dt}

Dividing the two heat flow equations gives:
dQ1dtdQ2dt=LfLv\frac{\frac{dQ_1}{dt}}{\frac{dQ_2}{dt}} = \frac{L_f}{L_v}

Substituting the expressions for dQ1dt\frac{dQ_1}{dt} and dQ2dt\frac{dQ_2}{dt}:
400KAλx300KA(10−λ)x=80540\frac{\frac{400KA}{\lambda x}}{\frac{300KA}{(10 - \lambda)x}} = \frac{80}{540}

400(10−λ)300λ=80540\frac{400(10 - \lambda)}{300\lambda} = \frac{80}{540}

4(10−λ)3λ=427\frac{4(10 - \lambda)}{3\lambda} = \frac{4}{27}

Dividing both sides by 44:
10−λ3λ=127\frac{10 - \lambda}{3\lambda} = \frac{1}{27}

10−λλ=327=19\frac{10 - \lambda}{\lambda} = \frac{3}{27} = \frac{1}{9}

Cross-multiplying:
9(10−λ)=λ9(10 - \lambda) = \lambda
90−9λ=λ90 - 9\lambda = \lambda
10λ=90  ⟹  λ=910\lambda = 90 \implies \lambda = 9

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