Do wanny z 40 litrami wody o temperaturze 40C, dolano 10 litrów wody o temperaturze 20C. Oblicz jaka będzie temperatura po wymieszaniu.

[latex]dane:\m_1 = 10 kg (1 l wody = 1 kg)\t_1 = 20^{o}C\m_2 = 40 kg\t_2 = 40^{o}C\szukane:\t_{k}=?[/latex] [latex]Q_{pbrane} = 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{10kg*20^{o}C+40kg*40^{o}C}{10kg+40kg}=frac{1800}{50} ^{o}C\\t_{k} = 36^{o}C[/latex]