(C) In a steady state,the rate of heat flow through both rods must be equal.Let the junction temperature be $\phi$.The rate of heat flow is given by $

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(C) In a steady state,the rate of heat flow through both rods must be equal.Let the junction temperature be $\phi$.The rate of heat flow is given by $H = \frac{KA(\Delta T)}{L}$.For the first rod: $H_1 = \frac{K A (\phi - 0)}{1}$.For the second rod: $H_2 = \frac{(3K) A (100 - \phi)}{2}$.Since $H_1 = H_2$:$\frac{KA\phi}{1} = \frac{3KA(100 - \phi)}{2}$$\phi = \frac{3}{2}(100 - \phi)$$2\phi = 300 - 3\phi$$5\phi = 300$$\phi = 60^{\circ}C$.

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

Difficult

Two rods have thermal conductivities $K$ and $3K$ and lengths $1 \ cm$ and $2 \ cm$ respectively. Their cross-sectional areas are equal. They are joined in series as shown in the figure. If the temperatures of the outer ends of this composite rod are $0^{\circ}C$ and $100^{\circ}C$ respectively,find the junction temperature $\phi$ in $^{\circ}C$.

A
$50$
B
$\frac{100}{3}$
C
$60$
D
$\frac{200}{3}$

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Class 11

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Chapter

10-2.Heat Transfer

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Heat Conduction and Thermal Conductivity

Consider two rods of same length and different specific heats $(S_{1}, S_{2})$,conductivities $(K_{1}, K_{2})$ and area of cross-sections $(A_{1}, A_{2})$ and both having temperatures $T_{1}$ and $T_{2}$ at their ends. If the rate of loss of heat due to conduction is equal,then:

MediumAIPMT 2002

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Two plates of equal area are placed in series. The ratios of their thicknesses and thermal conductivities are both $2:3$. The temperature of the outer surface of one plate is $100 ^\circ C$ and that of the other is $0 ^\circ C$. The temperature of the common surface is ....... $^\circ C$.

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The thickness of a metallic plate is $0.4 \ cm$. The temperature difference between its two surfaces is $20^{\circ}C$. The quantity of heat flowing per second is $50 \ \text{calories}$ through an area of $5 \ cm^2$. In the $CGS$ system,the coefficient of thermal conductivity is:

Medium

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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 $K$).

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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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Consider two rods of same length and different specific heats $(S_{1}, S_{2})$,conductivities $(K_{1}, K_{2})$ and area of cross-sections $(A_{1}, A_{2})$ and both having temperatures $T_{1}$ and $T_{2}$ at their ends. If the rate of loss of heat due to conduction is equal,then:

Two plates of equal area are placed in series. The ratios of their thicknesses and thermal conductivities are both $2:3$. The temperature of the outer surface of one plate is $100 ^\circ C$ and that of the other is $0 ^\circ C$. The temperature of the common surface is ....... $^\circ C$.

The thickness of a metallic plate is $0.4 \ cm$. The temperature difference between its two surfaces is $20^{\circ}C$. The quantity of heat flowing per second is $50 \ \text{calories}$ through an area of $5 \ cm^2$. In the $CGS$ system,the coefficient of thermal conductivity is:

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 $K$).

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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