In Lee's disc experiment two discs are separated by gap of thickness 5 mm. The space between the discs contains a gas of thermal conductivity 3.88 xx 10^(-5) "Wm"^(-1) "K"^(-1).At the steady state the temperature, of the two sides of discs are 368K and 333K. If the area of cross-section the slab is 25 "cm"^(2), calculate the quantity of heat crossing the gas per second.

In Lee's disc experiment two discs are separated by gap of thickness 5

1.	In Lee's disc experiment two discs are separated by gap of thickness 5 mm. The space between the discs contains a gas of thermal conductivity 3.88 xx 10^(-5) "Wm"^(-1) "K"^(-1).At the steady state the temperature, of the two sides of discs are 368K and 333K. If the area of cross-section the slab is 25 "cm"^(2), calculate the quantity of heat crossing the gas per second.
Answer» Solution :`x=5 xx 10^(-3) m, K= 3.88 xx 10^(-5) "Wm"^(-1) "K"^(-1), theta_1 - theta_2 = 368 - 33 = 35 "K"` `A = 25 "cm"^(2) = 25 xx 10^(4) "m"^(2) , t = 1 "Sec". Q= ?` `Q= "KA" ((theta_1 - theta_2) )/( x ) t = ( 3.88 xx 10^(-5) xx 25 xx 10^(-4) xx 35 xx 1)/( 5 xx 10^(-3) )` `= 6.79 xx 10^(-4)` J