Question

Tolong jawab no 18 pake caranya pleassse.... buat besok ini...mau di kumpulin...

Answer 1

Verified answer

K=Q/t .L/A.(X-35)
4k=Q/t.L/A.(165-X)

4/(X-35)=1/(165-X)
X=139

Answer 2

Verified answer

KALOR
• pindah panas

konduksi
k₁ = k. k₂ = 4k
A₁ = A₂ = A
L₁ = L₂
T₁ = 35 °C. T₂ = 175 °C
T = __?

perambatan kalor secara konduksi
H = Q / t = k A ∆T / L

pada keadaan setimbang
H₁ = H₂
k₁ A ∆T₁ / L = k₂ A ∆T₂ / L
k₁ (T - T₁) = k₂ (T₂ - T)
k (T - 35) = 4k (165 - T)
T - 35 = 4 (165 - T)
T - 35 = 660 - 4T
5T = 695
T = 695/5 = 139 °C ✓


atau
T = (k₁ T₁ + k₂ T₂) / (k₁ + k₂)
T = (k • 35 + 4k • 165) / (k + 4k)
T = (35k + 660k) / 5k
T = 695k / 5k = 139 °C ✓