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One end of a $0.25 m$ long metal bar is in steam and the other is in contact with ice. If $12 g$ of ice melts per minute, then the thermal conductivity of the metal is (Given cross section of the bar $= 5 \times 10^{- 4} m^{2}$ and latent heat of ice is $80 cal g^{- 1}$ ).

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$20 cal s^{- 1} m^{- 1} ° C^{- 1}$

$80 cal s^{- 1} m^{- 1} ° C^{- 1}$

$10 cal s^{- 1} m^{- 1} ° C^{- 1}$

$40 cal s^{- 1} m^{- 1} ° C^{- 1}$

answer is D.

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Detailed Solution

Here, $x = 0.25 m, T_{1} - T_{2} = 100 - 0 = 100 ° C$

$t = 1 \min = 60 s, A = 5 \times 10^{- 4} m^{2}, L = 80 cal g^{- 1}$

$Q = mL = 12 \times 80 = 960 cal$

But, $Q = \frac{KA (T_{1} - T_{2}) t}{x}$

or $960 = \frac{K \times 5 \times 10^{- 4} \times 100 \times 60}{0.25}$

$K = \frac{960 \times 0.25}{5 \times 10^{- 2} \times 60} = 80 cal s^{- 1} m^{- 1} ° C^{- 1}$

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