(B) The rate of heat transfer $H$ is given by $H = \frac{K A \Delta T}{l}$.Given: $A = 9.0 \, cm^2 = 9.0 \times 10^{-4} \, m^2$,$l = 0.54 \, m$,$\Delt

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(B) The rate of heat transfer $H$ is given by $H = \frac{K A \Delta T}{l}$.Given: $A = 9.0 \, cm^2 = 9.0 \times 10^{-4} \, m^2$,$l = 0.54 \, m$,$\Delta T = 100 \, ^\circ C - 0 \, ^\circ C = 100 \, K$.The rate of heat flow is also equal to the rate of melting of ice: $H = \frac{m L_f}{t}$,where $L_f = 334 \, J/g$ (latent heat of fusion of ice).$H = \frac{1 \, g \times 334 \, J/g}{33 \, s} \approx 10.12 \, W$.Equating the two expressions: $\frac{K \times 9.0 \times 10^{-4} \times 100}{0.54} = \frac{334}{33}$.$K \times \frac{0.09}{0.54} = 10.12$.$K \times \frac{1}{6} = 10.12$.$K = 60.72 \, W m^{-1} K^{-1}$.Rounding to the nearest integer,$K = 60 \, W m^{-1} K^{-1}$.

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

$A$ metallic rod of cross-sectional area $9.0 \, cm^2$ and length $0.54 \, m$,with the surface insulated to prevent heat loss,has one end immersed in boiling water and the other in an ice-water mixture. The heat conducted through the rod melts the ice at the rate of $1 \, g$ for every $33 \, s$. The thermal conductivity of the rod is ....... $W m^{-1} K^{-1}$.

A
$330$
B
$60$
C
$600$
D
$33$

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Class 11

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2 Thermal Properties of Matter: Heat Transfer

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Heat Conduction and Thermal Conductivity

Rate of flow of heat through a cylindrical rod is $H_1$. The temperatures at the ends of the rod are $T_1$ and $T_2$. If all the dimensions of the rod become double and the temperature difference remains the same,the rate of flow of heat becomes $H_2$. Then:

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$A$ cylindrical rod has temperatures $\theta_1$ and $\theta_2$ at its ends. The rate of heat flow is $Q$. All the linear dimensions of the rod are doubled while keeping the temperatures constant. The new rate of flow of heat is

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Two bars of thermal conductivities $K$ and $3K$ and lengths $1\, cm$ and $2\, cm$ respectively have equal cross-sectional area. They are joined length-wise as shown in the figure. If the temperatures at the ends of this composite bar are $0\,^{\circ}C$ and $100\,^{\circ}C$ respectively,then the temperature $\varphi$ of the interface is......... $^oC$.

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The area of the glass of a window of a room is $10\;m^2$ and thickness is $2\;mm$. The outer and inner temperatures are $40^{\circ}C$ and $20^{\circ}C$ respectively. The thermal conductivity of glass in the $MKS$ system is $0.2\;W/(m\cdot K)$. The heat flowing into the room per second will be:

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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 $^{\circ}C$.

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Rate of flow of heat through a cylindrical rod is $H_1$. The temperatures at the ends of the rod are $T_1$ and $T_2$. If all the dimensions of the rod become double and the temperature difference remains the same,the rate of flow of heat becomes $H_2$. Then:

$A$ cylindrical rod has temperatures $\theta_1$ and $\theta_2$ at its ends. The rate of heat flow is $Q$. All the linear dimensions of the rod are doubled while keeping the temperatures constant. The new rate of flow of heat is

The area of the glass of a window of a room is $10\;m^2$ and thickness is $2\;mm$. The outer and inner temperatures are $40^{\circ}C$ and $20^{\circ}C$ respectively. The thermal conductivity of glass in the $MKS$ system is $0.2\;W/(m\cdot K)$. The heat flowing into the room per second will be:

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