(A) The heat conducted through the ice layer of thickness $x$ in time $dt$ is given by $dQ = \frac{KA(T_2 - T_1)}{x} dt$.Here,$T_2 = 0^{\circ}C$ (temp

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(A) The heat conducted through the ice layer of thickness $x$ in time $dt$ is given by $dQ = \frac{KA(T_2 - T_1)}{x} dt$.Here,$T_2 = 0^{\circ}C$ (temperature at the water-ice interface) and $T_1 = -26^{\circ}C$ (outside air temperature).So,$dQ = \frac{KA(0 - (-26))}{x} dt = \frac{26KA}{x} dt$.This heat causes an additional thickness $dx$ of water to freeze. The mass of this new ice layer is $dm = A \cdot dx \cdot \rho$.The heat released during this phase change is $dQ = dm \cdot L = A \cdot dx \cdot \rho \cdot L$.Equating the two expressions for $dQ$:$\frac{26KA}{x} dt = A \cdot dx \cdot \rho \cdot L$.Rearranging to find the rate of increase of thickness $\frac{dx}{dt}$:$\frac{dx}{dt} = \frac{26K}{\rho x L}$.

Class 11 Physics 10-2.Heat Transfer Heat Conduction and Thermal Conductivity

Medium

$A$ deep rectangular pond of surface area $A$,containing water (density $=\rho$,specific heat capacity $=s$),is located in a region where the outside air temperature is at a steady value of $-26^{\circ}C$. The thickness of the frozen ice layer in this pond at a certain instant is $x$. Taking the thermal conductivity of ice as $K$ and its specific latent heat of fusion as $L$,the rate of increase of the thickness of the ice layer at this instant would be given by:

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A
$26 K / (\rho x L)$
B
$26 K / (\rho x^2 L)$
C
$26 K / (\rho L)$
D
$13 K / (\rho x L)$

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10-2.Heat Transfer

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Heat Conduction and Thermal Conductivity

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Three rods of same dimensions have thermal conductivities $3K, 2K$ and $K$. They are arranged as shown in the figure. In the steady state,the temperature of the junction $P$ is:

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Two plates $A$ and $B$ have thermal conductivities $84 \, W m^{-1} K^{-1}$ and $126 \, W m^{-1} K^{-1}$ respectively. They have the same surface area and same thickness. They are placed in contact along their surfaces. If the temperatures of the outer surfaces of $A$ and $B$ are kept at $100^{\circ} C$ and $0^{\circ} C$ respectively,then the temperature of the surface of contact in steady state is $.......... \, ^{\circ} C$.

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$A$ cylindrical rod with one end in a steam chamber and the other end in ice results in melting of $0.1 \ gm$ of ice per second. If the rod is replaced by another with half the length and double the radius of the first,and if the thermal conductivity of the material of the second rod is $\frac{1}{4}$ that of the first,the rate at which ice melts in $gm/sec$ will be:

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Two walls of thickness $d_1$ and $d_2$ and thermal conductivities $k_1$ and $k_2$ are in contact. In the steady state,the temperatures of the outer surfaces are $T_1$ and $T_2$. Find the temperature of the common interface.

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$A$ copper rod of length $10 \ cm$ and cross-sectional area $100 \ cm^2$ is to conduct $4000 \ J/s$ of heat. The thermal conductivity of copper is $400 \ W/m \cdot ^\circ C$. The temperature difference between the two ends of the rod must be maintained at ............. $^\circ C$.

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Three rods of same dimensions have thermal conductivities $3K, 2K$ and $K$. They are arranged as shown in the figure. In the steady state,the temperature of the junction $P$ is:

Two walls of thickness $d_1$ and $d_2$ and thermal conductivities $k_1$ and $k_2$ are in contact. In the steady state,the temperatures of the outer surfaces are $T_1$ and $T_2$. Find the temperature of the common interface.

$A$ copper rod of length $10 \ cm$ and cross-sectional area $100 \ cm^2$ is to conduct $4000 \ J/s$ of heat. The thermal conductivity of copper is $400 \ W/m \cdot ^\circ C$. The temperature difference between the two ends of the rod must be maintained at ............. $^\circ C$.

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