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:

  • A
    $\frac{k(T_1 - T_2)}{LA}$
  • B
    $kLA(T_1 - T_2)$
  • C
    $\frac{kA(T_1 - T_2)}{L}$
  • D
    $\frac{kL(T_1 - T_2)}{A}$

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Two plates of equal area are placed in contact with each other. Their thicknesses are $2.0 \ cm$ and $5.0 \ cm$. The temperature of the outer surface of the first plate is $-20^{\circ}C$ and the temperature of the outer surface of the second plate is $20^{\circ}C$. If the ratio of their thermal conductivities is $2:5$,find the temperature of the contact surface in $^{\circ}C$.

Two layers of a wall,$A$ and $B$,are made of different materials. Both layers have the same thickness. For layer $A$,the thermal conductivity is $K_A = 3 K_B$. The total temperature difference across the wall is $20^{\circ}C$. Find the temperature difference across layer $A$ in $^{\circ}C$.

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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In steady state heat conduction,the equations that determine the heat current $j(r)$ [heat flowing per unit time per unit area] and temperature $T(r)$ in space are exactly the same as those governing the electric field $E(r)$ and electrostatic potential $V(r)$ with the equivalence given in the table below.
Heat flow Electrostatics
$T(r)$ $V(r)$
$j(r)$ $E(r)$

We exploit this equivalence to predict the rate $\dot{Q}$ of total heat flowing by conduction from the surfaces of spheres of varying radii,all maintained at the same temperature. If $\dot{Q} \propto R^{n}$,where $R$ is the radius,then the value of $n$ is

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