One end of a copper rod of length 1.0 ; m and | vedclass

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(C) The heat transferred through the rod in time $t$ is given by the formula: $Q = \frac{KA(	heta_1 - 	heta_2)t}{l}$.This heat is used to melt the i

Vedclass · Accepted Answer

$6.9 	imes 10^{-3} \; kg$

Vedclass · Answer

(C) The heat transferred through the rod in time $t$ is given by the formula: $Q = \frac{KA(	heta_1 - 	heta_2)t}{l}$.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 of ice.Equating the two,we get: $mL = \frac{KA(	heta_1 - 	heta_2)t}{l}$.Given values: $K = 92 \; cal/(m \cdot s \cdot ^\circ C)$,$A = 10^{-3} \; m^2$,$l = 1.0 \; m$,$	heta_1 = 100^\circ C$,$	heta_2 = 0^\circ C$,$t = 60 \; s$,and $L = 8 	imes 10^4 \; cal/kg$.Substituting these values: $m = \frac{92 	imes 10^{-3} 	imes (100 - 0) 	imes 60}{1.0 	imes 8 	imes 10^4}$.$m = \frac{92 	imes 10^{-3} 	imes 100 	imes 60}{8 	imes 10^4} = \frac{92 	imes 6}{8 	imes 10^3} = \frac{552}{8000} = 0.069 	imes 10^{-1} = 6.9 	imes 10^{-3} \; kg$.

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