V = 1 l = 1 dm³ = 0,001 dm³ d = 1000 kg /m³ t₁ = 30 C t₂ = 60 C t = ? 1. obl. masę wody : d = m / V m = d * V m₁ = 1000kg/m³ * 0,02m³ = 20 kg m₂ = 1000 kg/m³ * 0,03 m³ = 30 kg 2. obl. t = temp. końcowa Q₁ = Q₂ m₁ * cw * ( t - t₁ ) = m₂ * cw * ( t₂ - t )/: cw m₁ * ( t - t₁ ) = m₂ * ( t₂ - t ) 20 kg * ( t - 30 C ) = 30 kg * (60 C - t ) 20 kg * t - 600 kg * C = 1800 kg * C - 30 kg * t 20 kg * t + 30 kg * t = 1800 kg * C + 600 kg * C 50 kg * t = 2400 kg * C / 50 kg t = 48 C
[latex]dane:\m_1 = 20 kg (1 l wody = 1 kg)\t_1 = 30^{o}C\m_2 = 30 kg\t_2 = 60^{o}C\szukane:\t_{k} = ?[/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_1t_{k}-m_1t_1 = m_2t_2-m_2t_{k}\\m_1t_{k}+m_2t_{k} = m_1t_1+m_2t_2\\t_{k}(m_1+m_2) = m_1t_1+m_2t_2}) /:(m_1+m_2)\\t_{k} = frac{m_1t_1+m_2t_2}{m_1+m_2}=frac{20kg*30^{o}C+30kg*60^{o}C}{20kg+30kg} = frac{2400}{50} ^{o}C\\t_{k} = 48^{o}C[/latex]
