Do 20 l wody o temperaturze pokojowej dolano 10 l wody o tem 60 stopni celsjusza .Oblicz temperature kąpieli . PROSZE DAJE NAJ !!!

[latex]dane:\m_1 = 20 kg (1 l wody = 1 kg)\t_1 = 20^{o}C\m_2 = 10 kg\t_2 = 60^{o}C\szukane:\t_{k}=?[/latex] [latex]Bilans energii wewnetrznej\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)[/latex] [latex]t_{k} = frac{m_1t_1+m_2t_2}{m_1+m_2}\\t_{k} = frac{20kg*20^{o}C+10kg*60^{o}C}{20kg+10kg} = 33,(3)^{o}C\\t_{k}approx33,3^{o}C}[/latex]