A conducting rod of length 1 ,m has an area o | vedclass

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(B) The rate of heat flow through the rod is given by the formula: $Q = \frac{KA(\Delta 	heta)t}{\ell}$.This heat is used to melt the ice,so $Q = mL$

Vedclass · Accepted Answer

$7.2 	imes 10^{-3} \,kg$

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(B) The rate of heat flow through the rod is given by the formula: $Q = \frac{KA(\Delta 	heta)t}{\ell}$.This heat is used to melt the ice,so $Q = mL$,where $m$ is the mass of ice melted and $L$ is the latent heat of fusion.Equating the two expressions: $mL = \frac{KA(\Delta 	heta)t}{\ell}$.Given values: $K = 96 \,cal/(s \cdot m \cdot ^{\circ}C)$,$A = 10^{-3} \,m^2$,$\Delta 	heta = 100^{\circ}C - 0^{\circ}C = 100^{\circ}C$,$t = 60 \,s$,$\ell = 1 \,m$,and $L = 8 	imes 10^4 \,cal/kg$.Substituting these values into the equation:$m = \frac{96 	imes 10^{-3} 	imes 100 	imes 60}{1 	imes 8 	imes 10^4}$.$m = \frac{96 	imes 10^{-1} 	imes 60}{8 	imes 10^4} = \frac{576}{8 	imes 10^5} = 72 	imes 10^{-4} \,kg = 7.2 	imes 10^{-3} \,kg$.

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