[tex]dane:\\t_{k} = 70^{o}C\\m_1 + m_2 = 200 \ g\\t_1 = 10^{o}C\\t_2 = 100^{o}C\\szukane:\\m_1, m_2 = ?\\\\Rozwiazanie\\\\Korzystamy \ z \ bilansu \ 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}) \ \ /:(t_2-t_{k})[/tex]
[tex]m_2 = m_1\cdot\frac{t_{k}-t_1}{t_2-t_{k}}\\\\m_2 = m_1\cdot\frac{70-10}{100-70}\\\\m_2 = m_1\cdot\frac{60}{30}\\\\m_2 = 2 \ m_1\\\\m_1 +m_2 = 200\\\\m_1 + 2m_1 = 200\\\\3m_1 = 200 \ \ /:3\\\\m_1 = \frac{200}{3} = 66\frac{2}{3} \ g\\\\m_2 = 200 - 66\frac{2}{3} = 133\frac{1}{3} \ g[/tex]
Odp. Należy zmieszać 66 i 2/3 g wody o temperaturze 10°C i 133 i 1/3 g wody o temperaturze 100°C.
