An ice cube of dimensions 60,cm times 50,cm t | vedclass

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Question

(B) The rate of heat flow is given by $\frac{dQ}{dt} = \frac{KA\Delta T}{\ell}$.The surface area $A$ of the box is $2(0.6 	imes 0.5 + 0.5 	imes 0.2

Vedclass · Accepted Answer

$61 	imes 10^{-5}\,kg\,s^{-1}$

Vedclass · Answer

(B) The rate of heat flow is given by $\frac{dQ}{dt} = \frac{KA\Delta T}{\ell}$.The surface area $A$ of the box is $2(0.6 	imes 0.5 + 0.5 	imes 0.2 + 0.2 	imes 0.6) = 2(0.3 + 0.1 + 0.12) = 2(0.52) = 1.04\,m^2$.Given thermal conductivity $K = 0.05\,W\,m^{-1\circ}C^{-1}$,thickness $\ell = 1\,cm = 0.01\,m$,and temperature difference $\Delta T = 40^{\circ}C - 0^{\circ}C = 40^{\circ}C$.Substituting the values:$\frac{dQ}{dt} = \frac{0.05 	imes 1.04 	imes 40}{0.01} = 0.05 	imes 1.04 	imes 4000 = 208\,J/s$.The rate of melting of ice $m$ is given by $\frac{dQ}{dt} = m L_f$,where $L_f = 3.4 	imes 10^5\,J/kg$.$m = \frac{208}{3.4 	imes 10^5} = \frac{208}{3.4} 	imes 10^{-5} \approx 61.17 	imes 10^{-5}\,kg/s$.Thus,the rate of melting is approximately $61 	imes 10^{-5}\,kg/s$.

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