A metal rod of length 10 text{ cm} and area o | vedclass

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(D) Given:Length of rod,$l = 10 	ext{ cm} = 0.1 	ext{ m}$Area of cross-section,$A = 2.8 	imes 10^{-4} 	ext{ m}^2$Temperature difference,$\Delta T

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$100$

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(D) Given:Length of rod,$l = 10 	ext{ cm} = 0.1 	ext{ m}$Area of cross-section,$A = 2.8 	imes 10^{-4} 	ext{ m}^2$Temperature difference,$\Delta T = 80^{\circ} 	ext{C} - 0^{\circ} 	ext{C} = 80 	ext{ K}$Mass of melted ice,$m = 20 	ext{ g}$Time taken,$t = 5 	ext{ min} = 300 	ext{ s}$Latent heat of ice,$L = 80 	ext{ cal/g} = 80 	imes 4.184 	ext{ J/g} = 334.72 	ext{ J/g}$Heat required to melt ice,$Q = m 	imes L = 20 	ext{ g} 	imes 334.72 	ext{ J/g} = 6694.4 	ext{ J}$Rate of heat flow,$H = \frac{Q}{t} = \frac{6694.4 	ext{ J}}{300 	ext{ s}} \approx 22.314 	ext{ W}$Using the formula for thermal conduction,$H = \frac{k A \Delta T}{l}$$22.314 = \frac{k 	imes (2.8 	imes 10^{-4} 	ext{ m}^2) 	imes 80 	ext{ K}}{0.1 	ext{ m}}$$22.314 = k 	imes 0.224$$k = \frac{22.314}{0.224} \approx 99.61 	ext{ J s}^{-1} 	ext{ m}^{-1} 	ext{ K}^{-1}$Rounding to the nearest integer,$k \approx 100 	ext{ J s}^{-1} 	ext{ m}^{-1} 	ext{ K}^{-1}$.

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