A cubic metal block of mass 5 ,kg and edge le | vedclass

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(B) The rate of heat loss from the block is given by $dQ/dt = -hA(T - T_{surr})$,where $h$ is the heat transfer coefficient,$A$ is the surface area,an

Vedclass · Accepted Answer

$1000$

Vedclass · Answer

(B) The rate of heat loss from the block is given by $dQ/dt = -hA(T - T_{surr})$,where $h$ is the heat transfer coefficient,$A$ is the surface area,and $T_{surr}$ is the ambient temperature.Since $dQ = mc dT$,we have $mc(dT/dt) = -hA(T - T_{surr})$.Here,$m = 5 \,kg$,$c = 500 \,J/kg/^{\circ}C$,$h = 50 \,W/m^2/^{\circ}C$,and $T_{surr} = 0^{\circ}C$.The surface area $A$ of the cube exposed to air consists of $5$ faces (since one face is on an insulating surface): $A = 5 	imes (0.1 \,m)^2 = 5 	imes 0.01 = 0.05 \,m^2$.Substituting the values: $5 	imes 500 	imes (dT/dt) = -50 	imes 0.05 	imes (T - 0)$.$2500 	imes (dT/dt) = -2.5 	imes T$.$dT/T = -(2.5 / 2500) dt = -0.001 dt$.Integrating from $T_i = 100^{\circ}C$ to $T_f = 37^{\circ}C$:$\int_{100}^{37} (dT/T) = \int_{0}^{t} -0.001 dt$.$\ln(37/100) = -0.001 t$.$\ln(0.37) \approx -0.994$.$-0.994 = -0.001 t \implies t = 994 \,s$.The value closest to $994 \,s$ is $1000 \,s$.

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