As per the question,
Thickness of the pane=0.4cm
Area=2×104cm2
Temperature difference between the pan (ΔT)=25∘C
Let the conductivity of the material =K
Hence, rate of the heat flow dtdQ=lKAdθ
Now, substituting the values,
⇒dtdQ=0.4cmK×25×2×106cm2=125×K×106J/sec
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