(D) The heat produced by the heater per unit time is $P = \frac{V^2}{R} = \frac{200^2}{20} = 2000 \ W = 2000 \ J/s$.The heat conducted through the gla

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(D) The heat produced by the heater per unit time is $P = \frac{V^2}{R} = \frac{200^2}{20} = 2000 \ W = 2000 \ J/s$.The heat conducted through the glass window per unit time is given by $H = \frac{K A (T_{in} - T_{out})}{d}$.Here, $K = 0.2 \ W/(m \cdot K)$, $A = 1 \ m^2$, $d = 0.2 \ cm = 0.002 \ m$, and $T_{in} = 20^{\circ}C$.Equating the heat produced to the heat conducted: $2000 = \frac{0.2 \times 1 \times (20 - T_{out})}{0.002}$.$2000 = 100 \times (20 - T_{out})$.$20 = 20 - T_{out}$.$T_{out} = 0^{\circ}C$. Since this result is not among the options, the correct choice is $D$.

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

Medium

$A$ room is maintained at $20^{\circ}C$ by a heater of resistance $20 \ \Omega$ connected to $200 \ V$ mains. The temperature is uniform throughout the room and heat is transmitted through a glass window of area $1 \ m^2$ and thickness $0.2 \ cm$. What will be the temperature outside in $^{\circ}C$? Given that thermal conductivity $K = 0.2 \ J/(s \cdot m \cdot K)$ for glass and $J = 4.2 \ J/cal$.

IIT 1978 All IIT

A
$15.24$
B
$15.00$
C
$24.15$
D
None of the above

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10-2.Heat Transfer

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Heat Conduction and Thermal Conductivity

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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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If the thermal conductivity of copper is $9$ times that of steel,what will be the temperature of the junction in $^oC$?

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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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