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given that
high h=1.18 m
wide b=2.19 m
thickness L=1.448 cm=1.448*10^-2 m
higher temperature T2=32 c
lower temperature T1=24 c
co efficient of thermal conductivity K=0.81wm^-1 C^-1
area of the glass A=1.18*2.19=2.5842 m^2
this problem belongs to principle of conduction
according to heat produced in conduction
thermal coductivity = Q=kA(T2-T1) t/l
where Q= heat flow through the glass
k=co.efficient of thermal conductivity
A=area of glass
(T2-T1) = temperature difference
t= time
L=thickness of glass
so the rate of heat follow =P=Q/t = kA(T2-T1)/L
=0.81*2.5842*(32-24)/1.448*10^-2
=1156.5
=1156.5 watts
therefore the rate of heat transfer P=1156.5 watts
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