[tex]dane:\\m_1 = m_2 = m = 0,5 \ kg\\Q_1 = Q_2 = Q = 20 \ kJ = 20 \ 000 \ J\\c_1 = 4200\frac{J}{kg\cdot^{o}C} \ - cieplo \ wlasciwe \ wody\\c_2 = 1640\frac{J}{kg\cdot^{o}C} \ - cieplo \ wlasciwe \ oleju\\szukane:\\\Delta T_1 =?\\\Delta T_2 = ?\\\\Rozwiazanie\\\\Q = m\cdot c\cdot \Delta T \ \ /:(m\cdot c)\\\\\Delta T} = \frac{Q}{m\cdot c}\\\\\\\Delta T_1 = \frac{Q}{m\cdot c_1}\\\\\Delta T_1 = \frac{20 \ 000 \ J}{0,5 \ kg\cdot4200\frac{J}{kg\dot^{o}C}}}\\\\\underline{\Delta T_1 \approx9,5^{o}C}[/tex]
[tex]\Delta T_2 = \frac{Q}{m\cdot c_2}\\\\\Delta T_2 = \frac{20 \ 000 \ J}{0,5 \ kg\cdot1640\frac{J}{kg\cdot^{o}C}}\\\\\underline{\Delta T_2 \approx24,4^{o}C}[/tex]
