[tex]dane:\\P = 2,5 \ kW = 2 \ 500 \ W\\m = 2 \ kg\\T_1 = 25^{o}C\\T_2 = 100^{o}C\\\Delta T = T_2 - T_1 = 100^{o}C - 25^{o}C = 75^{o}C\\c = 4200\frac{J}{kg\cdot^{o}C}\\szukane:\\t = ?\\\\Rozwiazanie\\\\P = \frac{W}{t} \ \ /\cdot t\\\\W = P\cdot t\\oraz\\W = Q = c\cdot m\cdot \Delta T\\\\P\cdot t = c\cdot m\cdot \Delta T \ \ /:P\\\\t = \frac{c\cdot m\cdot \Delta T}{P}\\\\t = \frac{4200\frac{J}{kg\cdot^{o}C}\cdot2 \ kg\cdot75^{o}C}{2500 \ W}=\frac{630000 \ J}{2500 \ W} = \frac{630000 \ W\cdot s}{2500 \ W}[/tex]
[tex]t = 252 \ s = 4 \ min \ i \ 12 \ s[/tex]
Odp. Czajnik musi być włączony przez 252 sekundy.
