i = 0 T(l = 1OCPC- Boundary Condition
1^2
Ąt° - 2T° +t})
^ = 20 + 0.4239(20-2(20) + 100) = 20 + 0.4239(80)
= 20 + 33.912 = 53.912°C
r2‘ = t° +a.{t? -2T° +t°)
= 20 + 0.4239(20 - 2(20) + 20)
= 20 + 0.4239(0)
= 20 + 0 = 20°C
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