[tex]dane:\\m = 2 \ kg \ \ \ (1 \ litr \ wody = 1 \ kg)\\T_1 = 20^{o}C\\T_2 = 100^{o}C\\\Delta T = T_2 - T_1 = 100^{o}C - 20^{o}C = 80^{o}C\\c = 4190\frac{J}{kg\cdot^{o}C} \ - \ cieplo \ wlasciwe \ wody\\P = 2,2 \ kW \\U = 230 \ V\\t = 6 \ min = 6\cdot60 \ s = 360 \ s\\\eta = 60\%\\szukane:\\R = ?\\\\Rozwiazanie\\\\\eta = \frac{P_{uzyteczna}}{P{calkowita}} = \frac{W_{uzyteczna}}{W_{calkowita}}\\\\W_{uz} = m\cdot c\cdot \Delta T\\\\W_{cal} = U\cdot I\cdot t[/tex]
[tex]\eta = \frac{m\cdot c\cdot \Delta T}{U\cdot I\cdot t}\\\\\eta\cdot U\cdot I\cdot t = m\cdot c\cdot \Delta T \ \ /:(\eta\cdot U\cdot t})\\\\I = \frac{m\cdot c\cdot \Delta T}{\eta\cdot U\cdot t} = \frac{2 \ kg\cdot4190\frac{J}{kg\cdot^{o}C}\cdot80^{o}C}{0,6\cdot230 \ V\cdot360^{o}} = 13,5 \ A\\\\R = \frac{U}{I}\\\\R = \frac{230 \ V}{13,5 \ A} = 17 \ \Omega[/tex]
Odp. Opór grzałki czajnika elektrycznego R = 17 Ω.
