[tex]Dane:\\m_1 = m_2 = m = 2 \ kg\\\Delta T_1=20^{o}C\\c_1 = 2420\frac{J}{kg\cdot^{o}C} \ - \ cieplo \ wlasciwe \ gliceryny\\c_2 = 2090\frac{J}{kg\cdot^{o}C} \ - \ cieplo \ wlasciwe \ benzyny\\Q_1 = Q_2\\Szukane:\\\Delta T_2 = ?\\\\Q_1 = m\cdot c_1 \cdot \Delta T_1\\\\Q_2 = m\cdot c_2\cdot\Delta T\\\\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\\\\\Delta T_2 = \frac{2430}{2090}\cdot20^{o}C[/tex]
[tex]\boxed{\Delta T_2 \approx 23,3 ^{o}C}[/tex]
Odp. Temperatura beznyny wzrosłaby o ok. 23,3°C.
