[tex]Dane:\\U = 220 \ V\\t = 2 \ min = 2\cdot60 \ s = 120 \ s\\m = 1 \ kg \ \ (1 \ l \ wody = 1 \ kg)\\T_1 = 30^{o}C\\T_2 = 100^{o}C\\\DElta T = T_2 - T_1 = 100^{o}C - 30^{o}C = 70^{o}C\\c = 4200\frac{J}{kg\cdot^{o}C}\\Szukane:\\I = ?\\\\Rozwiazanie\\\\E = W = U\cdot I \cdot t\\oraz\\E = Q = m\cdot c\cdot \Delta T\\\\U\cdot I\cdot t = m\cdot c\cdot \Delta T \ \ /:(U\cdot t)[/tex]
[tex]I = \frac{m\cdot c\cdot \Delta T}{U\cdot t}\\\\I = \frac{1 \ kg\cdot 4200\frac{J}{kg\cdot^{o}C}\cdot70^{o}C}{220 \ V \cdot 120 \ s}=\frac{294000 \ J}{26400 \ V\cdot s}\\\\\underline{I \approx11,14 \ A}[/tex]