[tex]dane:\\T_1 = 5^{o}C\\T_2 = 25^{o}C\\\Delta T = T_2 - T_1 = 25^{o}C - 5^{o}C = 20^{o}C\\Q = 2,8 \ kJ = 2 \ 800 \ J\\c = 130\frac{J}{kg\cdot^{o}C} \ - \ cieplo \ wlasciwe \ rteci\\szukane:\\m = ?\\\\Q = c\cdot m \cdot \Delta T \ \ /:(c\cdot \Delta T)\\\\m = \frac{Q}{c\cdot \Delta T}\\\\m = \frac{2800 \ J}{130\frac{J}{kg\cdot^{o}C}\cdot20^{o}C}\approx 1,077 \ kg[/tex]
