[tex]dane:\\Q = 23,1 \ kJ = 23 \ 100 \ J\\T_1 = 10^{o}C\\T_2 = 65^{o}C\\\DElta T = T_2 - T_1 = 65^{o}C - 10^{o}C = 55^{o}C\\c = 4200\frac{J}{kg\cdot^{o}C}\\szukane:\\m = ?\\\\Rozwiazanie\\\\Q = c\cdot m\cdot\Delta T \ \ /:(c\cdot \Delta T)\\\\m = \frac{Q}{c\cdot \Delta T}\\\\m = \frac{23100}{4200\cdot55} \ [kg]\\\\m = 0,1 \ kg[/tex]