A conducting rod of length 1 ,m has an area o | vedclass

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(B) The rate of heat flow through the rod is given by the formula: $Q = \frac{KA(\Delta 	heta)t}{\ell}$.This heat is used to melt the ice,so $Q = mL$

Vedclass · Accepted Answer

$7.2 	imes 10^{-3} \,kg$

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(B) The rate of heat flow through the rod is given by the formula: $Q = \frac{KA(\Delta 	heta)t}{\ell}$.This heat is used to melt the ice,so $Q = mL$,where $m$ is the mass of ice melted and $L$ is the latent heat of fusion.Equating the two expressions: $mL = \frac{KA(\Delta 	heta)t}{\ell}$.Given values: $K = 96 \,cal/(s \cdot m \cdot ^{\circ}C)$,$A = 10^{-3} \,m^2$,$\Delta 	heta = 100^{\circ}C - 0^{\circ}C = 100^{\circ}C$,$t = 60 \,s$,$\ell = 1 \,m$,and $L = 8 	imes 10^4 \,cal/kg$.Substituting these values into the equation:$m = \frac{96 	imes 10^{-3} 	imes 100 	imes 60}{1 	imes 8 	imes 10^4}$.$m = \frac{96 	imes 10^{-1} 	imes 60}{8 	imes 10^4} = \frac{576}{8 	imes 10^5} = 72 	imes 10^{-4} \,kg = 7.2 	imes 10^{-3} \,kg$.

Similar Questions

The two ends of a rod of length $L$ and a uniform cross-sectional area $A$ are kept at two temperatures $T_1$ and $T_2$ $(T_1 > T_2)$. The rate of heat transfer,$\frac{dQ}{dt}$,through the rod in a steady state is given by:

The rate of heat flow through the cross-section of the rod shown in the figure is ($T_2 > T_1$ and the thermal conductivity of the material of the rod is $K$).

How is thermal conduction represented for a given temperature difference for any material by the rate of heat conduction?

For a spherical shell with inner and outer radii $r_1$ and $r_2$ respectively,the formula for the heat flow rate is = ......... where $T_1$ and $T_2$ are the temperatures of the inner and outer surfaces of the shell $(T_1 > T_2)$. The thermal conductivity of the material of the shell is $k$.

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