A window used to thermally insulate a room from outside consists of two parallel glass sheets each of area 2.6 `m^(2)` and thickness 1 cm separated by 5 cm thick stagnant air. In the steady state, the room glass interface is at `18^(@)`C and the glass-outdoor interface is at and air are respectively 0.8`Wm^(-1) K^(-1)` and 0.08 `Wm^(-1) K^(-1)`, the rate of flow of heat through the window is

Options

A. 15 W
B. 40 W
C. 60 W
D. 80 W

Answer

D

Explanation

According to the question, room-glass interface and glass-outdoor interface are connected in series combination so, the equivalent
thermal resistance,
`R_(eq) = R_(1) + R_(2) + R_(3)`
`therefore " " R_(1) = (l_(1))/(k_(1)A_(1)), R_(2) = (l_(2))/(k_(2)A_(2)), R_(3) = (l_(3))/(k_(3) A_(3))`
Where, `k_(1) and k_(3)` are thermal conductivity of glass and `k_(2)` is thermal conductivity of air.
or ` " " R_(eq) = (l_(1))/(k_(1) A_(1)) + (l_(2))/(k_(2)A_(2)) + (l_(3))/(k_(3)A_(3))`
Given, `l_(1) = l_(3) = 1 cm = 10^(-2)` m,
`l_(2) = 5 cm = 5 xx 10^(-2) `m
`k_(1) = k_(3) = 0.8 Wm^(-1) K^(-1)`
`k_(2) = 0.08 Wm^(-1) K^(-1)` ,
and `A_(1) = A_(2) = A_(3) = 2.6 m^(2)`
Putting the given values, we get
`(10^(2))/(0.8 xx 2.6) + (5 xx 10^(-2))/(0.08 xx 2.6 ) + (10^(-2))/(0.8 xx 2.6)`
`= ((10^(-2))/(2.6)) ((52)/(0.8)) = (1)/(4)`
hence, flow of heat,
` H =(Delta T)/(R_(eq)) = ([18 - (-2)])/(((1)/(4)))`
H = 80 W