A slab of stone of area 0.36, m^2 and thickne | vedclass

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(C) The heat required to melt the ice is given by $Q = m L_f$.The heat conducted through the slab is given by $Q = \frac{K A (T_1 - T_2) t}{L}$.Equati

Vedclass · Accepted Answer

$1.24$

Vedclass · Answer

(C) The heat required to melt the ice is given by $Q = m L_f$.The heat conducted through the slab is given by $Q = \frac{K A (T_1 - T_2) t}{L}$.Equating the two, we get $\frac{K A (T_1 - T_2) t}{L} = m L_f$.Rearranging for thermal conductivity $K$, we have $K = \frac{m L_f L}{A (T_1 - T_2) t}$.Given values: $m = 4.8\, kg$, $L_f = 3.36 	imes 10^5\, J/kg$, $L = 0.1\, m$, $A = 0.36\, m^2$, $(T_1 - T_2) = 100^{\circ} C$, and $t = 1\, hour = 3600\, s$.Substituting the values: $K = \frac{4.8 	imes 3.36 	imes 10^5 	imes 0.1}{0.36 	imes 100 	imes 3600}$.$K = \frac{4.8 	imes 3.36 	imes 10^4}{0.36 	imes 3.6 	imes 10^5} = \frac{16.128}{129.6} 	imes 10 = 1.24\, J/m/s/^{\circ} C$.

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