[tex]dane:\\t_1 = 0^{o}C\\m_2 = 20 \ g\\t_2 = 20^{o}C\\t_{k} = 2^{o}C\\C_1 = 4190\frac{J}{kg\cdot^{o}C} \ - \ cieplo \ wlasciwe \ wody\\C_2 = 2100\frac{J}{kg\cdt^{o}C} \ - \ cieplo \ wlasciwe \ nafty\\szukane:\\m_1 = ?\\\\Rozwiazanie\\\\Q_{pobrane} = Q_{oddane}\\\\m_1C_1(t_{k}-t_1) = m_2C_2(t_2-t_{k}) \ \ /:C_1(t_{k}-t_1)\\\\m_1 = \frac{m_2C_2(t_2-t_{k})}{C_1(t_{k}-t_1)}[/tex]
[tex]m_1 = \frac{20 \ g\cdot2100\frac{J}{kg\cdot^{o}C}\cdot(20^{o}-2^{o}C)}{4190\frac{J}{kg\cdot^{o}C}\cdot(2^{o}C - 0^{o}C)}\\\\m_1 \approx90 \ g[/tex]
Odp. Trzeba dodać ok. 90 g zimnej wody.
