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Question: One end of a 0.25 m long metal bar is in steam and the other is in contact with ice. If 2 g of ice m...

One end of a 0.25 m long metal bar is in steam and the other is in contact with ice. If 2 g of ice melts per minute, then the thermal conductivity of the metal is (Given cross section of the bar =5×10−4m2= 5 \times 10^{- 4}m^{2}and latent heat of ice is 80calg−180calg^{- 1})

A

20 cal s−1 m−1 ∘C−120\text{ cal }\text{s}^{- 1}\text{ }\text{m}^{- 1}\ {^\circ}C^{- 1}

B

10 cal s−1 m−1 ∘C−110\text{ cal }\text{s}^{- 1}\text{ }\text{m}^{- 1}\ {^\circ}C^{- 1}

C

40 cal s−1 m−1 ∘C−140\text{ cal }\text{s}^{- 1}\text{ }\text{m}^{- 1}\ {^\circ}C^{- 1}

D

80 cal s−1 m−1 ∘C−180\text{ cal }\text{s}^{- 1}\text{ }\text{m}^{- 1}\ {^\circ}C^{- 1}

Answer

80 cal s−1 m−1 ∘C−180\text{ cal }\text{s}^{- 1}\text{ }\text{m}^{- 1}\ {^\circ}C^{- 1}

Explanation

Solution

Here, x=0.25m,T1−T2=100−0=100∘Cx = 0.25m,T_{1} - T_{2} = 100 - 0 = 100{^\circ}C

T = 1 min = 60s, A = 5 × 10−4m2,10^{- 4}m^{2}, L = 80 calg−1g^{- 1}

But, Q=KA(T1−T2)txQ = \frac{KA(T_{1} - T_{2})t}{x}

Or 960=K×5×10−4×100×600.25960 = \frac{K \times 5 \times 10^{- 4} \times 100 \times 60}{0.25}

∴K=960×0.2550×10−2×60\therefore K = \frac{960 \times 0.25}{50 \times 10^{- 2} \times 60}

=80cals−1m−1∘C−1= 80cals^{- 1}m^{- 1}{^\circ}C^{- 1}