The only possibility of heat flow in a thermo | vedclass

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Question

(A) The heat required to melt $500 \ g$ of ice at $0^\circ C$ is given by $Q = mL$,where $m = 500 \ g$ and $L = 80 \ cal/g$.$Q = 500 	imes 80 = 40,00

Vedclass · Accepted Answer

$2.47$

Vedclass · Answer

(A) The heat required to melt $500 \ g$ of ice at $0^\circ C$ is given by $Q = mL$,where $m = 500 \ g$ and $L = 80 \ cal/g$.$Q = 500 	imes 80 = 40,000 \ cal$.The rate of heat flow through the cork is given by $\frac{Q}{t} = \frac{KA\Delta	heta}{\Delta x}$,where $K = 0.0075 \ cal/(cm \cdot s \cdot ^\circ C)$,$A = 75 \ cm^2$,$\Delta	heta = (40 - 0) = 40^\circ C$,and $\Delta x = 5 \ cm$.Substituting the values,we get:$\frac{40,000}{t} = \frac{0.0075 	imes 75 	imes 40}{5}$$\frac{40,000}{t} = 0.0075 	imes 15 	imes 40$$\frac{40,000}{t} = 4.5$$t = \frac{40,000}{4.5} \approx 8888.89 \ s$.To convert the time into hours,divide by $3600$:$t = \frac{8888.89}{3600} \approx 2.47 \ hr$.

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