[tex]dane:\\m_1 = m_2 = m = 3 \ kg\\\Delta T_1 = 10^{o}C\\c_1 = 4200\frac{J}{kg\cdot^{o}C}\\c_2 = 2200\frac{J}{kg\cdot^{o}C}\\Q_1 = Q_2\\szukane:\\\Delta T_2 = ?\\\\\\Q = m\cdot c\cdot \Delta T\\\\Q_1 = m\cdot c_1\cdot \Delta T_1\\\\Q_2 = m\cdot c_2\cdot \Delta T_2\\\\Q_1 = Q_2\\\\m\cdot c_1\cdot \Delta T_1 = m\cdot c_2\cdot \Delta T_2 \ \ /:(m\cdot c_2)\\\\\Delta T_2 = \frac{c_1}{c_2}\cdot \Delta T_1}[/tex]
[tex]\Delta T_2 = \frac{4200\frac{J}{kg\cdot^{o}C}}{2200\frac{J}{kg\cdot^{o}C}}\cdot10^{o}C\\\\\underline{\Delta T_2 \approx 19,1^{o}C}[/tex]
Odp. Temperatura nafty wzrosłaby o ≈ 19,1 °C.
