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Standard XII

Physics

Thermal Resistance

A wall of thi...

Question

A wall of thickness $0.6 m$ width has a normal area $1.5 m^{2}$ and is made up of a material of thermal conductivity $0.4 W/mK$ . The temperature difference between the two sides is $800^{\circ} C .$ . Find the thermal resistance of the wall.

1 W/K

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1.8 W/K

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1 K/W

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1.8 K/W

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Solution

The correct option is C $1 K/W$

Thermal resistance $R = \frac{Δ x}{K A}$
Here, $Δ x = Width of the wall = 0.6 m$
$A = Normal area of wall = 1.5 m^{2}$
& $K = thermal conductivity = 0.4 W/mK$
$\Rightarrow R = \frac{0.6}{0.4 \times 1.5} = 1 K/W$ .
Thus, option (c) is the correct answer.

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Thermal Resistance

Standard XII Physics

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A wall of thickness 0.6 m width has a normal area 1.5 m2 and is made up of a material of thermal conductivity 0.4 W/mK. The temperature difference between the two sides is 800∘C.. Find the thermal resistance of the wall.

The correct option is C 1 K/W Thermal resistance R=ΔxKA Here, Δx=Width of the wall =0.6 m A=Normal area of wall =1.5 m2 & K=thermal conductivity =0.4 W/mK ⇒R=0.60.4×1.5=1 K/W. Thus, option (c) is the correct answer.

A wall of thickness $0.6 m$ width has a normal area $1.5 m^{2}$ and is made up of a material of thermal conductivity $0.4 W/mK$ . The temperature difference between the two sides is $800^{\circ} C .$ . Find the thermal resistance of the wall.

The correct option is C $1 K/W$

Thermal resistance $R = \frac{Δ x}{K A}$
Here, $Δ x = Width of the wall = 0.6 m$
$A = Normal area of wall = 1.5 m^{2}$
& $K = thermal conductivity = 0.4 W/mK$
$\Rightarrow R = \frac{0.6}{0.4 \times 1.5} = 1 K/W$ .
Thus, option (c) is the correct answer.