Petit soucis sur l'exercice n°1...

m₁c(Θ-Θ₁) = m₂c(Θ₂-Θ)
m₁cΘ+m₂cΘ = m₂cΘ₂+m₁cΘ₁
Θ = (m₂cΘ₂+m₁cΘ₁)/c(m₁+m₂)       Θ₁=273,15+10=283,15 K    Θ₂ =273,15+55=328,15 K  
Θ = (0,060x4,19x328,15+0,100x4,19x283,15)/4,19(0,060+0,100)
Θ = 300 K
Θ = 300-273,15
Θ = 27°C