[tex]dane:\\m_1 = 120 \ kg \ \ (1 \ l \ wody = 1 \ kg)\\t_1 = 20^{o}C\\t_2 = 100^{o}C\\t_{k} = 35^{o}C\\szukane:\\m_2 = ?\\\\Rozwiazanie\\\\Q_{pobrane} = Q_{oddane}\\\\m_1c(t_{k}-t_1) = m_2c(t_2-t_{k}) \ \ /:c(t_2-t_{k})\\\\m_2 = \frac{m_1(t_{k}-t_1)}{t_2-t_{k}}\\\\m_2 = \frac{120 \ kg\cdot(35^{o}C-20^{o}C)}{100^{o}C - 35^{o}C}=\frac{1800}{65} \ kg\\\\m_2 \approx27,69 \ kg \approx27,69 \ l[/tex]
Odp. Należy dolać ≈ 27,69 litra wrzątku.
