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Tb = 5C
kb = 2ka
Tc = ..............?
Penyelesaian :
k₁ A₁ ΔT₁/L₁ = k₂ A₂ ΔT₂/L₂
k₁ ΔT₁ = k₂ ΔT₂
ka • ΔTa = kb • ΔTb
ka • (80 - Tc) = 2ka • (Tc - 5)
1 • (80 - Tc) = 2 • (Tc - 5)
80 - Tc = 2Tc - 10
90 = 3Tc
Tc = 90/3
Tc = 30⁰C
Jadi, suhu pada batang kedua sambungan adalah 30⁰C