[tex]dane:\\m = 2 \ kg\\T_1 = 18^{o}C\\Q = 9,6 \ kJ = 9 \ 600 \ J\\c = 130\frac{J}{kg\cdot ^{o}C}} \ - cieplo \ wlasciwe \ olowiu\\szukane:\\T_2 = ?\\\\Rozwiazanie\\\\Q = m\cdot c\cdot(T_2-T_1) \ \ /:(m\cdot c)\\\\T_2-T_1 = \frac{Q}{m\cdot c}[/tex]
[tex]T_2 = T_1 + \frac{Q}{m\cdot c}\\\\T_2 = 18^{o}C + \frac{9600 \ J}{2 \ kg \cdot 130\frac{J}{kg\cdot^{o}C}}\\\\T_2 = 18^{o}C + 36,2^{o}C\\\\\underline{T_2 = 54,9^{o}C}[/tex]
Odp. Temperatura końcowa T₂ = 54,9 °C.
