Quis. [Fisika]
Q = m × c × ∆T
Q = 3 kg × 3.200J/KgC × (100 - 30)
Q = ...
Q = ...![]()
Q = m × c × ∆T
Q = 3 kg × 3.200J/KgC × (100 - 30)
Q = ...
Q = ...
Q = m.c.delta T
Q = 3 . 3.200 . 70
Q = 67200 J
Q = 67.2 kJ
maaf banget kalau salah :DD
cara 1 lebih simple =
Q = m.c.∆T
Q = 1kg x 4.200j/kg°c x (100-10)
Q = 4,2 x 90
Q = 378
cara 2 lengkap =
massa benda/air = m = 1kg
perubahan suhu/kenaikan suhu = ∆T
kalori jenis benda = C = 4.200 j/kg°c
Tanya? Besar Kalori/ kalori yang di resap = Q
Q = m.c.∆T
Q = 1 x 4.200 x 90 =
Q = 378
[answer.2.content]
