[tex]Dane:\\P = 1,5 \ kW = 1 \ 500 \ W = 1 \ 500 \ \frac{J}{s}\\T_1 = 20^{o}C\\T_2 = 50^{o}C\\\Delta T = T_2-T_1 = 50^{o}C - 20^{o}C = 30^{o}C\\t = 3 \ min = 3\cdot60 \ s = 180 \ s\\c = 4200\frac{J}{kg\cdot^{o}C}} \ - \ cieplo \ wlasciwe \ wody\\Szukane:\\m = ?\\\\Rozwiazanie\\\\W = P\cdot t\\oraz\\W = Q = m\cdot c\cdot \Delta T\\\\m\cdot c\cdot \Delta T = P\cdot t \ \ /:(c\cdot \Delta T)\\\\m = \frac{P\cdot t}{c\cdot \Delta T}\\\\m = \frac{1500\frac{J}{s}\cdot180 \ s}{4200\frac{J}{kg\cdot^{o}C}\cdot30^{o}}[/tex]
[tex]\boxed{m = 2,14 \ kg}[/tex]
