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Standard XII

Physics

Conduction Law

A room has a ...

Question

A room has a window fitted with a single 1.0 m $\times$ 2.0 m glass of thickness 2 mm. (a) Calculate the rate of heat flow through the closed window when the temperature inside the room is $32^{\circ} C$ and that outside is $40^{\circ} C$ . (b) The glass is now replaced by two glasspanes, each having a thickness of 1 mm and separated by a distance of 1 mm. Calculate the rate of heat flow under the same cnditions of temperature. Thermal conductivity of window glass = $1.0 J s^{- 1} m^{- 1 \circ} C^{- 1}$ and that of air = $0.025 J s^{- 1} m^{- 1 \circ} C^{- 1}$ .

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Solution

$\frac{Q}{t} = \frac{K A (θ_{1} - θ_{2})}{1}$

= $\frac{1 \times 2 (40 - 32)}{2 \times 10^{- 3}}$

= 8000 J/sec.

(b) Resistance of glass = $\frac{1}{a k_{g}} + \frac{1}{a k_{g}}$

Resistance of air = $\frac{1}{a k_{a}}$

Net resistance = $\frac{1}{a k_{g}} + \frac{1}{a k_{g}} + \frac{1}{a k_{g}}$

$(\frac{2}{k_{g}} + \frac{1}{k_{a}})$

$= \frac{1}{a} (\frac{2 k_{a} + k g}{k_{g} k_{a}})$

$= \frac{1 \times 10^{- 3}}{2} (\frac{2 \times 0.025 + 1}{0.025})$

$= \frac{1 \times 10^{- 3} \times 1.05}{0.05}$

$\frac{Q}{t} = \frac{(θ_{1} - θ_{2})}{R}$

= 385.9 = 381 W.

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Qt=KA(θ1−θ2)1 = 1×2(40−32)2×10−3 = 8000 J/sec. (b) Resistance of glass = 1akg+1akg Resistance of air = 1aka Net resistance = 1akg+1akg+1akg (2kg+1ka) =1a(2ka+kgkgka) =1×10−32(2×0.025+10.025) =1×10−3×1.050.05 Qt=(θ1−θ2)R = 385.9 = 381 W.