(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 \ c

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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 2 Thermal Properties of Matter: 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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2 Thermal Properties of Matter: Heat Transfer

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Heat Conduction and Thermal Conductivity

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There is a layer of snow $x \ cm$ thick on water,when the temperature of the air is $-\theta ^\circ C$ (less than the freezing point). If the thickness of the layer increases from $x$ to $y$ in time $t$,then the value of $t$ is given by:

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