[tex]dane:\\m = 2 \ kg \ \ (1 \ l \ wody = 1 \ kg)\\T_1 = 15^{o}C\\T_2 = 100^{o}C\\\Delta T = T_2 - T_1 = 100^{o}C - 15^{o}C = 85^{o}C\\P = 2,5 \ kW = 2 \ 500 \ W\\c = 4200\frac{J}{kg\cdot^{o}C}\\szukane:\\t = ?\\\\Rozwiazanie\\\\P = \frac{W}{t} \ \ \rightarrow \ \ t = \frac{W}{P}\\\\W = Q = c\cdot m\cdot \Delta T = 4200\frac{J}{kg\cdot^{o}C}\cdot2 \ kg \cdot85^{o}C = 714 \ 000 \ J\\\\t = \frac{W}{P} = \frac{714000 \ J}{2500 \ W}=285,6 \ s = 4,76 \ minuty[/tex]
