Length of the square plate is at Ti=0∘C is L0=15cm=0.15m ,thus A0=L02=0.0225m2 ,Tf=400∘C
αL=1.1×10−5K−1 .
Let,A the required area and ΔT=Tf−Ti=400K .
We know that ,
A=A0(1+2αLΔT)
Thus on plugin the values in above equation we get,
A=0.0225(1+2×1.1×10−5×400)⟹A=0.022698m2
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