A slab of stone of area 3600 , cm^2 and thick | vedclass

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(D) Given: Area of slab $(A) = 3600 \, cm^2 = 0.36 \, m^2$.Thickness $(d) = 10 \, cm = 0.1 \, m$.Temperature difference $(\Delta 	heta) = 100^{\circ}

Vedclass · Accepted Answer

$1.24$

Vedclass · Answer

(D) Given: Area of slab $(A) = 3600 \, cm^2 = 0.36 \, m^2$.Thickness $(d) = 10 \, cm = 0.1 \, m$.Temperature difference $(\Delta 	heta) = 100^{\circ} C - 0^{\circ} C = 100 \, K$.Time $(t) = 1 \, hour = 3600 \, s$.Mass of ice melted $(m) = 4.8 \, kg$.Latent heat of fusion $(L) = 3.36 	imes 10^5 \, J/kg$.The heat required to melt the ice is $Q = m 	imes L = 4.8 	imes 3.36 	imes 10^5 \, J$.The rate of heat flow is given by $\frac{Q}{t} = \frac{K A \Delta 	heta}{d}$.Substituting the values:$\frac{4.8 	imes 3.36 	imes 10^5}{3600} = \frac{K 	imes 0.36 	imes 100}{0.1}$.$\frac{1612800}{3600} = K 	imes 360$.$448 = K 	imes 360$.$K = \frac{448}{360} \approx 1.24 \, J \, s^{-1} \, m^{-1} \, K^{-1}$.

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