Jadi suhu T1 = 55°C dan T2 = 43,75°C
Dik
k p = 1/4 kq
k r = 3 kq
Dit T1 dan T2?
Lihat pada batang PQ
To p = 100°C
Toq = T2
T sambungan = T1
T1 > T2
Q/t p = Q/t q
kp × A × ΔTp /L = kq x A x ΔTq/L
krn ukuran sama
kp x ΔTp = kq x ΔTq
¹/₄ kq x (100 - T1) = kq x (T1 - T2)
coret kq pd kedua ruas
¹/₄ x (100 - T1) = T1 - T2
100 - T1 = 4 (T1 - T2)
100 - T1 = 4T1 - 4T2
100 = 5T1 - 4T2
5T1 - 4T2 = 100
Lihat batang QR
To q =T1
To r = 40ºC
T sambungan = T2
Q/t q = Q/t r
kq × A × ΔTq /L = kr x A x ΔTr/L
kq x ΔTq = kr x ΔTr
kq x (T1 - T2) = 3kq x (T2 - 40)
T1 - T2 = 3 (T2 - 40)
T1 - T2 = 3T2 - 120
T1 - 4T2 = - 120
Cari T1 dan T2
T1 - 4 T2 = - 120
_____________ _
4T1 = 220
T1 = 220 ÷ 4 = 55°C
T1 - 4T2 = -120
T1 + 120 = 4 T2
4T2 = 55 + 120 = 175
T2 = 43,75°
Kategorisasi
Kelas : XI
Mapel : Fisika
Materi : Suhu dan Kalor
Kode Kategorisasi : 11.6.5.
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Jadi suhu T1 = 55°C dan T2 = 43,75°C
Dik
k p = 1/4 kq
k r = 3 kq
Dit T1 dan T2?
Lihat pada batang PQ
To p = 100°C
Toq = T2
T sambungan = T1
T1 > T2
Q/t p = Q/t q
kp × A × ΔTp /L = kq x A x ΔTq/L
krn ukuran sama
kp x ΔTp = kq x ΔTq
¹/₄ kq x (100 - T1) = kq x (T1 - T2)
coret kq pd kedua ruas
¹/₄ x (100 - T1) = T1 - T2
100 - T1 = 4 (T1 - T2)
100 - T1 = 4T1 - 4T2
100 = 5T1 - 4T2
5T1 - 4T2 = 100
Lihat batang QR
To q =T1
To r = 40ºC
T sambungan = T2
T1 > T2
Q/t q = Q/t r
kq × A × ΔTq /L = kr x A x ΔTr/L
krn ukuran sama
kq x ΔTq = kr x ΔTr
kq x (T1 - T2) = 3kq x (T2 - 40)
coret kq pd kedua ruas
T1 - T2 = 3 (T2 - 40)
T1 - T2 = 3T2 - 120
T1 - 4T2 = - 120
Cari T1 dan T2
5T1 - 4T2 = 100
T1 - 4 T2 = - 120
_____________ _
4T1 = 220
T1 = 220 ÷ 4 = 55°C
T1 - 4T2 = -120
T1 + 120 = 4 T2
4T2 = 55 + 120 = 175
T2 = 43,75°
Kategorisasi
Kelas : XI
Mapel : Fisika
Materi : Suhu dan Kalor
Kode Kategorisasi : 11.6.5.