Air teh panas bersuhu80 derajatC melepaskan kalor sebanyak 12,6J.Massa air teh panas tersebut 500 gram.jika kalor jenisnya 4200J/KgderajatC,suhu setelah terjadi

T1 = 80°C
Q = 12,6 KJ = 12600 J
m = 500 gram
c = 4,2 J/gram°C
T2 = ...?

Q = m x c x ∆T
Q = m x c x (T2 - T1)
12600 = 500 x 4,2 x (T2 - 80)
12600 = 2100 x (T2 - 80)
T2 - 80 = 12600/2100
T2 - 80 = 6
T2 = 6 + 80
T2 = 86°C