[tex]dane:\\m_1 = 135 \ kg \ \ (1 \ l \ wody = 1 \ kg)\\t_1 = 22^{o}C\\t_2 = 100^{o}C\\t_{k} = 36^{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{135 \ kg\cdot(36^{o}C - 22^{o}C)}{100^{o}C - 36^{o}C} = \frac{135 \ kg\cdot14^{o}C}{64^{o}C}\\\\m_2 \approx29,53 \ kg = 29,53 \ l[/tex]
Odp. Należy dolać ok. 29,53 l wrzątku.
