[tex]dane:\\m_1 = 6 \ kg\\t_1 = 20^{o}C\\m_2 = 2 \ kg\\t_2 = 100^{o}C\\t_{k} = 23,3^{o}C\\c_1 = 4186\frac{J}{kg\cdot^{o}C} \ - \ cieplo \ wlasciwe \ wody\\szukane:\\c_2=? \ - \cieplo \ wlasciwe \ zelaza\\\\Rozwiazanie\\\\Q_{pobrane} = Q_{oddane}\\\\m_1c_1(t_{k}-t_1) = m_2c_2(t_2-t_{k}) \ \ /:c_2(t_2-t_{k})\\\\c_2 = c_1\cdot\frac{m_1(t_{k}-t_1)}{m_2(t_2-t_{k})}[/tex]
[tex]c_2 = 4186\frac{J}{kg\cdot^{o}C}\cdot\frac{6 \ kg(23,3^{o}C - 20^{o}C)}{2 \ kg(100^{o}C - 23,3^{o}C)}\\\\\underline{c_2 = 540\frac{J}{kg\cdot^{o}C}}[/tex]
