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प्रश्न
A room has a window fitted with a single 1.0 m × 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°C and the outside is 40°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 conditions of temperature. Thermal conductivity of window glass = 1.0 J s−1 m−1°C−1 and that of air = 0.025 m-1°C-1 .
उत्तर
(a) 
Length, l = 2 mm = 0.0002 m
Rate of flow of heat `=(Delta"T")/(l/("KA"))`
`={1xx2xx(4032)}/{2xx10^-3}`
`={KA.DeltaT}/{l} `
`= {1xx2xx(40-32)}/{2xx10^-3}`
= 8000 J / sec
(b) 
|
Resistance of glass, `R_g = {l}/{K_g.A}`
Resistance of air, `R_A = {l}/{K_A.A}`
From the circuit diagram, we can find that all the resistors are connected
`R_s = R_g + R_A + R_g`
`=10^-3/2 (2/K_g + 1/K_A) `
`=10^-3 ( 2/1 + 1/0.025)`
`= 10^-3/2 xx((2xx0.025 + 1))/0.025`
Rate of flow of heat ,= `q ={DeltaT}/{R_5}`
`= ( T_1 - T_2)/R_s`
`= {(40-32) xx2xx0.025}/{40^_3xx(2xx0.025 + 1 )}`
=381W
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