Dane Szukane Wzór zimna woda m₁ = 5 l = 5 kg Q₁ = m·cw·ΔT₁ t₁ = 10 ⁰C t₃ = 18 ⁰C cw = 4200 J/kg·⁰C ΔT₁ = (t₃ - t₁) Ciepła woda m₂ = 10 l = 10 kg t₂ = ? Q₂ = m·cw·ΔT₂ t₃ = 18 ⁰C cw = 4200 J/kg·⁰C ΔT₂ = (t₃ - t₂) Rozwiązanie z bilansu cieplnego Q₁ = Q₂ m₁·cw·ΔT₁ = m₂·cw·ΔT₂ /:cw m₁·ΔT₁ = m₂·ΔT₂ m₁·(t₃ - t₁) = m₂·(t₂ - t₃) m₁(t₃ - t₁) = m₂t₂ - m₂t₃ m₂t₂ = m₂t₃ + m₁(t₃ - t₁) t₂ = m₂t₃ + m₁(t₃ - t₁) / m₂ rachunek jednostek t = [ kg·⁰C + kg·⁰C/kg ] = [kg·⁰C/kg] = [ ⁰C ] t₂ = 10·18 + 5·(18 - 10)/10 t = 180 + 40/10 t = 220/10 t = 22 ⁰C Odp. Temperatura początkowa ciepłej wody wynosi 22 ⁰C
q[latex]dane:\1 l wody = 1 kg\m_1 = 5 kg\t_1 = 10^{o}C\m_2 = 10 kg\t_{k} = 18^{o}C\szukane:\t_2 = ?[/latex] [latex]Q_{pobrane} = Q_{oddane}\m_1C(t_{k}-t_1) = m_2C(t_2-t_{k}) /:C\\m_1(t_{k}-t_1) = m_2(t_2-t_{k})\\m_1(t_{k}-t_1) = m_2t_2 - m_2t_{k}\\m_2t_2 = m_1(t_{k}-t_1)+m_2t_{k} /:m_2\\t_2 = frac{m_1(t_{k}-t_1)+m_2t_{k}}{m_2}=frac{5kg(18^{o}C-10^{o}C)+10kgcdot18^{o}C}{10kg}= frac{220}{10}^{o}C\\t_2 = 22^{o}C[/latex]
