[tex]dane:\\I = 2 \ A\\U = 220 \ V\\m = 1 \ kg\\\Delta T = 20^{o}C\\c = 400\frac{J}{kg\cdot^{o}C}\\szukane:\\t = ?\\\\Rozwiazamie\\\\W = U\cdot I\cdot t\\oraz\\W = Q = c\cdot m\cdot \Delta T\\\\U\cdot I\cdot t = c\cdot m\cdot \Delta T \ \ /:(U\cdot I)\\\\t = \frac{c\cdot m\cdot\Delta T}{U\cdot I}\\\\t = \frac{400\frac{J}{kg\cdot^{o}C}\cdot1 \ kg\cdot20^{o}C}{220V\cdot2 \ A}\\\\t =\frac{400\cdot 20 \ J}{220\frac{kg\cdot m^{2}}{A\cdot s^{3}}\cdot 2 \ A}[/tex]
[tex]t = \frac{4000 \ kg\cdot\frac{m^{2}}{s^{2}}}{220 \ kg\cdot\frac{m^{2}}{s^{3}}}\\\\t \approx18 \ s[/tex]
Odp. Czas przepływu prądu wynosił ok. 18 s.
