(B) Given: Thickness $\Delta x = 5 \ mm = 5 \times 10^{-3} \ m$, Area $A = 5 \ cm^2 = 5 \times 10^{-4} \ m^2$, Temperature difference $\Delta T = 14^{

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(B) Given: Thickness $\Delta x = 5 \ mm = 5 \times 10^{-3} \ m$, Area $A = 5 \ cm^2 = 5 \times 10^{-4} \ m^2$, Temperature difference $\Delta T = 14^{\circ} C$, Heat flow rate $Q = 42 \ J/s = 42 \ W$.The formula for heat conduction in steady state is $Q = \frac{K A \Delta T}{\Delta x}$.Rearranging for thermal conductivity $K$: $K = \frac{Q \Delta x}{A \Delta T}$.Substituting the values: $K = \frac{42 \times 5 \times 10^{-3}}{5 \times 10^{-4} \times 14}$.Simplifying the expression: $K = \frac{42 \times 10^{-3}}{10^{-4} \times 14} = \frac{3 \times 10^{-3}}{10^{-4}} = 3 \times 10 = 30 \ W \ m^{-1} \ K^{-1}$.

Class 11 Physics 2 Thermal Properties of Matter: Heat Transfer Heat Conduction and Thermal Conductivity

Easy

The thickness of a uniform rectangular metal plate is $5 \ mm$ and the area of each surface is $5 \ cm^2$. In steady state, the temperature difference between the two surfaces of the plate is $14^{\circ} C$. If the heat flowing through the plate in one second from one surface to the other surface is $42 \ J$, then the thermal conductivity of the metal is

TS EAMCET 2024 All TS EAMCET

A
$90 \ W \ m^{-1} \ K^{-1}$
B
$30 \ W \ m^{-1} \ K^{-1}$
C
$45 \ W \ m^{-1} \ K^{-1}$
D
$60 \ W \ m^{-1} \ K^{-1}$

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2 Thermal Properties of Matter: Heat Transfer

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Heat Conduction and Thermal Conductivity

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Two metal rods $1$ and $2$ have the same length and the same temperature difference at their ends. Their thermal conductivities are $K_1$ and $K_2$ and their cross-sectional areas are $A_1$ and $A_2$ respectively. The condition for the same rate of heat flow in them is ........

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$A$ rod of length $L$ with sides fully insulated is made of a material whose thermal conductivity varies with temperature $T$ as $K = \frac{\alpha}{T}$,where $\alpha$ is a constant. The ends of the rod are kept at temperatures $T_1$ and $T_2$. The temperature $T$ at a distance $x$ from the end kept at $T_1$ is:

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$A$ cylindrical metallic rod in thermal contact with two heat reservoirs at its two ends conducts an amount of heat '$Q_1$' in time '$t$'. The metallic rod is melted and the material is formed into a rod of length four times the length of the original rod. The amount of heat conducted by the new rod when placed in thermal contact with the same two reservoirs in time '$t$' is '$Q_2$'. Then $\frac{Q_1}{Q_2}$ is:

MediumMHT CET 2022

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Three rods of equal length and cross-sectional area with thermal conductivities $K, 2K$,and $3K$ are joined as shown in the figure. The temperatures of their outer ends are $110\ ^oC, 20\ ^oC$,and $0\ ^oC$ respectively. Find the temperature of the junction in $^oC$.

Medium

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Two sheets of thickness $d$ and $2d$ and the same area are touching each other on their faces. The temperatures $T_A$,$T_B$,and $T_C$ shown are in a geometric progression with a common ratio $r = 2$. Then,the ratio of the thermal conductivity of the thinner sheet to the thicker sheet is:

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Two metal rods $1$ and $2$ have the same length and the same temperature difference at their ends. Their thermal conductivities are $K_1$ and $K_2$ and their cross-sectional areas are $A_1$ and $A_2$ respectively. The condition for the same rate of heat flow in them is ........

Three rods of equal length and cross-sectional area with thermal conductivities $K, 2K$,and $3K$ are joined as shown in the figure. The temperatures of their outer ends are $110\ ^oC, 20\ ^oC$,and $0\ ^oC$ respectively. Find the temperature of the junction in $^oC$.

Two sheets of thickness $d$ and $2d$ and the same area are touching each other on their faces. The temperatures $T_A$,$T_B$,and $T_C$ shown are in a geometric progression with a common ratio $r = 2$. Then,the ratio of the thermal conductivity of the thinner sheet to the thicker sheet is:

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