Tolongin kak,+caranya pliss Jika 75 gram air yang suhu nya 0°C dicampur dengan 50 gram air yang suhunya 100°C,maka suhu air campuran tersebut adalah.... (kalor jenis air = 1,00 kal/g°C)
DB45
Misal suhu akhir = T m1 = 75 g t1 = 0 C Δt1 = T - 0 = T
m2 = 50 g t2 = 100 C Δt2 = t1 - T = 100 - T c air = 1 kal/g.C
kalor lepas = kalor terima m1 c.air Δt1 = m2. c. air . Δt2 --> c. air sama m1. Δt1 = m2 Δt2 75(T) = 50(100- T) 75 T = 5.000 - 50T 125 T = 5.000 T = 5.000/125 T = 40° C suhu air campuranb = 40°C
m1 = 75 g
t1 = 0 C
Δt1 = T - 0 = T
m2 = 50 g
t2 = 100 C
Δt2 = t1 - T = 100 - T
c air = 1 kal/g.C
kalor lepas = kalor terima
m1 c.air Δt1 = m2. c. air . Δt2 --> c. air sama
m1. Δt1 = m2 Δt2
75(T) = 50(100- T)
75 T = 5.000 - 50T
125 T = 5.000
T = 5.000/125
T = 40° C
suhu air campuranb = 40°C