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Standard VII

Physics

Conduction Law

One end of a ...

Question

One end of a uniform brass rod 20 cm long and $10 c m^{2}$ cross-sectional area is kept at $100^{\circ} C$ . The other end is in perfect thermal contact with another rod of identical cross-section and length 10 cm. The free end of this rod is kept in melting ice and when the steady state has been reached, it is found that 360g of ice melts per hour. Calculate the thermal conductivity of the rod, given that the thermal conductivity of brass is $0.25 c a l / s c m^{\circ} C$ and L = 80 cal/g.

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Solution

$\frac{1}{K_{e q}} = \frac{1}{K_{1}} + \frac{1}{K_{2}} \Rightarrow \frac{1}{K_{e q}} = \frac{K_{1} K_{2}}{K_{1} + K_{2}} = \frac{0.25 x}{0.25 + x} Q = \frac{100}{\frac{30}{(\frac{0.25 x}{0.25 + x}) \times 10}} = \frac{360}{80} \Rightarrow \frac{25 x}{0.75 x + 3 x} = \frac{360}{80} \Rightarrow 25 x \times 8 = 36 (0.75 x + 3 x) \Rightarrow 200 x = 27 + 108 x \Rightarrow x = \frac{25}{92} = 0.293 c a l / s c m^{o} C$

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[L_{v} = 540 cal/g, L_{f} = 80 cal/ g]

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1Keq=1K1+1K2⇒1Keq=K1K2K1+K2=0.25x0.25+xQ=10030(0.25x0.25+x)×10=36080⇒25x0.75x+3x=36080⇒25x×8=36(0.75x+3x)⇒200x=27+108x⇒x=2592=0.293cal/scmoC