200 gram air bersuhu 80derajat celcius dicampurkab dengan 300 gram air bersuhu 20 derajat celcius, berapah suhu campurannya

Q lepas = Q terima
m1 x c1 x delta T1 = m2 x c2 x delta T2 -> c1 = c2 karna sama" air.
(200/1000) x (80 - Tc) = (300/1000) x (Tc - 20)
0,2 (80 - Tc) = 0,3 (Tc - 20)
80 - Tc = 0,3/0,5 (Tc - 20)
80 - Tc = 1,5 (Tc - 20)
80 - Tc = 1,5 Tc - 30
80 + 30 = 1,5 Tc + 1 Tc
110 = 2,5 Tc
Tc = 110/2,5
Tc = 44 derajat celcius

semoga membantu:)