One end of copper rod of uniform cross-section and of length 1-45 m is in contact with ice at 0- Cand the other end with water at 100- C- The position of point along its length where a temperature of 200 - C should be maintained so that in steady state the mass of ice melting is equal to that of steam produced in the same interval of time is x cm from hotter end of rod- Find x -Assume that the whole system is insulated from surroundings-

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Standard XII

Physics

Thermal Resistance

One end of co...

Question

One end of copper rod of uniform cross-section and of length $1.45 m$ is in contact with ice at $0^{\circ} C$ and the other end with water at $100^{\circ} C$ . The position of point along its length where a temperature of $200^{\circ} C$ should be maintained so that in steady state the mass of ice melting is equal to that of steam produced in the same interval of time is $x c m$ from hotter end of rod. Find $x$ [Assume that the whole system is insulated from surroundings].

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Solution

Let the x be distance of temperature 200 from boiling end
For boiling , $Q_{1} = k A (200 - 100) / x$
for ice $Q_{2} = k A (200 - 0) / (1.5 - x)$
we know rate of ice melting =steam produced
$\frac{Q_{1}}{L v} = \frac{Q_{2}}{L i}$ ,where $L_{i} = 80 c a l / g, L_{v} = 540 c a l / g$
so we get $\frac{k A (200 - 100)}{L_{v} x} = \frac{k A (200 - 0)}{L_{i} (1.5 - x)}$
$x = 10 \Rightarrow x - 1 = 9 c m$

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Let the x be distance of temperature 200 from boiling endFor boiling , Q1=kA(200−100)/xfor ice Q2=kA(200−0)/(1.5−x)we know rate of ice melting =steam producedQ1Lv=Q2Li ,where Li=80cal/g ,Lv=540cal/gso we get kA(200−100)Lvx=kA(200−0)Li(1.5−x)x=10⇒x−1=9cm