(C) The rate of heat flow $H$ through the rod is given by the formula: $H = \frac{KA(T_1 - T_2)}{L}$.Given: $K = 90 \ W \ m^{-1} \ K^{-1}$,$A = 4 \ cm

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(C) The rate of heat flow $H$ through the rod is given by the formula: $H = \frac{KA(T_1 - T_2)}{L}$.Given: $K = 90 \ W \ m^{-1} \ K^{-1}$,$A = 4 \ cm^2 = 4 \times 10^{-4} \ m^2$,$L = 20 \ cm = 0.2 \ m$,$T_1 = 100^{\circ} C$,$T_2 = 0^{\circ} C$.Substituting the values: $H = \frac{90 \times 4 \times 10^{-4} \times (100 - 0)}{0.2} = \frac{90 \times 4 \times 10^{-4} \times 100}{0.2} = \frac{3.6}{0.2} = 18 \ J/s$.Total heat $Q$ transferred in time $t = 7 \ minutes = 7 \times 60 = 420 \ s$ is: $Q = H \times t = 18 \times 420 = 7560 \ J$.The mass of ice melted $m$ is given by $Q = mL_f$,where $L_f = 336 \times 10^3 \ J/kg$.$m = \frac{Q}{L_f} = \frac{7560}{336 \times 10^3} = 0.0225 \ kg = 22.5 \ g$.

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

Medium

The length of a metal rod is $20 \ cm$ and its area of cross-section is $4 \ cm^2$. If one end of the rod is kept at a temperature of $100^{\circ} C$ and the other end is kept in ice at $0^{\circ} C$,then the mass of the ice melted in $7 \ minutes$ is (Thermal conductivity of the metal $= 90 \ W \ m^{-1} \ K^{-1}$ and latent heat of fusion of ice $= 336 \times 10^3 \ J \ kg^{-1}$) (in $g$)

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A
$20$
B
$67.5$
C
$22.5$
D
$45$

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10-2.Heat Transfer

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Heat Conduction and Thermal Conductivity

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The end $A$ of a rod $AB$ of length $1\,m$ is maintained at $80\,^{\circ}C$ and the end $B$ at $0\,^{\circ}C$. The temperature at a distance of $60\,cm$ from the end $A$ is......... $^{\circ}C$

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The coefficients of thermal conductivity of copper,mercury,and glass are respectively $K_c, K_m$,and $K_g$ such that $K_c > K_m > K_g$. If the same quantity of heat is to flow per second per unit area of each and corresponding temperature gradients are $X_c, X_m$,and $X_g$,then:

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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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$1.56 \times 10^5 \ J$ of heat is conducted through a $2 \ m^2$ wall of $12 \ cm$ thickness in one hour. The temperature difference between the two sides of the wall is $20^{\circ} C$. The thermal conductivity of the material of the wall is (in $W \ m^{-1} \ K^{-1}$):

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$A$ rod $A$ of length $40\, cm$ has a temperature difference of $80^\circ C$ at its two ends. Another rod $B$ of length $60\, cm$ has a temperature difference of $90^\circ C$ at its ends. Both rods have the same area of cross-section. If the rate of flow of heat is the same for both,then the ratio of their thermal conductivities $(K_A : K_B)$ will be:

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The end $A$ of a rod $AB$ of length $1\,m$ is maintained at $80\,^{\circ}C$ and the end $B$ at $0\,^{\circ}C$. The temperature at a distance of $60\,cm$ from the end $A$ is......... $^{\circ}C$

The coefficients of thermal conductivity of copper,mercury,and glass are respectively $K_c, K_m$,and $K_g$ such that $K_c > K_m > K_g$. If the same quantity of heat is to flow per second per unit area of each and corresponding temperature gradients are $X_c, X_m$,and $X_g$,then:

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:

$1.56 \times 10^5 \ J$ of heat is conducted through a $2 \ m^2$ wall of $12 \ cm$ thickness in one hour. The temperature difference between the two sides of the wall is $20^{\circ} C$. The thermal conductivity of the material of the wall is (in $W \ m^{-1} \ K^{-1}$):

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