dla wody 1 l = 1 dm³ ma masę 1 kg m₁ = n kg t₁ = 20 C m₂ = 2n kg t₂ = 80 C t = ? Q₁ = Q₂ m₁ * cw *Δt₁ = m₂ * cw *Δt₂ /: cw m₁ *Δt₁ = m₂ *Δt₂ nkg * (t - 20 C )= 2n kg * ( 80 - t ) / :n 1 kg ( t - 20 C ) = 2 kg * (80 - t ) 1 kg * t - 20 kg * C = 160 kg * C - 2 kg * t 1 kg * t + 2 kg * t = 160 kg * C + 20 kg * C 3 kg * t = 180 kg * C t = 180 kg * C /3 kg t = 60 C
Dane Szukane Wzór m₁ = n kg t₃ = ? Q = m·cw·ΔT m₂ = 2 n kg t₁ = 20 C t₂ = 80 C Rozwiązanie Q₁ = m₁·cw·ΔT₁ Q₂ = m₂·cw·ΔT₂ Q₁ = Q₂ m₁·cw·ΔT = m₂·cw·ΔT₂ /: cw m₁·ΔT₁ = m₂·ΔT₂ ΔT₁ = t₃ - t₁ ΔT₂ = t₂ - t₃ m₁(t₃ - t₁) = m₂(t₂ - t₃) m₁t₃ - m₁t₁ = m₂t₂ - m₂t₃ m₁t₃ + m₂t₃ = m₂t₂ + m₁t₁ t₃(m₁ + m₂) = m₂t₂ + m₁t₁ t₃ = m₂t₂ + m₁t₁/m₁ + m₂ t₃ = 2 n·80 + n·20/ n + 2 n t₃ = 160 n + 20 n /3 n t₃ = 180 n/3 n Rachunek jednostek t₃ = [ kg· ⁰C + kg· ⁰C/kg ] = [ kg·⁰C/kg] = [⁰C] t₃ = 60 ⁰C Odp. Temperatura końcowa mieszaniny wynosi 60 ⁰C
