The two opposite faces of a cubical piece of | vedclass

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(b) $Q = mL = KA\frac{{({	heta _1} - {	heta _2})}}{l}t$ $ \Rightarrow \;m = \frac{1}{L} 	imes KA\frac{{({	heta _1} - {	heta _2})}}{l} 	imes t$

Vedclass · Accepted Answer

$300 $

Vedclass · Answer

(b) $Q = mL = KA\frac{{({	heta _1} - {	heta _2})}}{l}t$ $ \Rightarrow \;m = \frac{1}{L} 	imes KA\frac{{({	heta _1} - {	heta _2})}}{l} 	imes t$

$ = \frac{1}{{80}} 	imes 0.2 	imes 4 	imes \frac{{(100 - 0)}}{{\sqrt 4 }} 	imes 10 	imes 60$

$ = \frac{{0.2 	imes 4 	imes 100 	imes 600}}{{80 	imes 2}} = 300\;gm$

The two opposite faces of a cubical piece of iron (thermal conductivity $= 0.2\, CGS$ units) are at ${100^o}C$ and ${0^o}C$ in ice. If the area of a surface is $4c{m^2}$, then the mass of ice melted in $10$ minutes will be ...... $gm$

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