[tex]dane:\\m = 2 \ kg\\T_1 = 60^{o}C\\T_2 = 100^{o}C\\\Delta T = T_2-T_1 = 100^{o}C - 60^{o}C = 40^{o}C\\t = 4 \ min = 4\cdot60 \ s = 240 \ s\\c = 4200\frac{J}{kg\cdot^{o}C} \ - \ cieplo \ wlasciwe \ wody\\szukane:\\P = ?\\\\Rozwiazanie\\\\W = Q = c\cdot m\cdot \Delta T\\\\W = 4200\frac{J}{kg\cdot^{o}C}\cdot2 \ kg\cdot40^{o}C = 336 \ 000 \ J\\\\P = \frac{W}{t}\\\\P = \frac{336000 \ J}{240 \ s} = 1 \ 400 \ W = 1,4 \ kW[/tex]
Odp. Moc tego urządzenia P = 1,4 kW.
