[tex]Dane:\\P = 1 \ 200 \ W\\m = 0,5 \ kg \ \ (1 \ l \ 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_{w} = 4200\frac{J}{kg\cdot^{o}C} \ - \ cieplo \ wlasciwe \ wody\\Szukane:\\t = ?\\\\Rozwiazanie\\\\P = \frac{W}{t} \ \ \ |\cdot t\\\\W = P\cdot t\\oraz\\W = Q = m\cdot c_{w}\cdot \Delta T\\\\P\cdot t = m\cdot c_{w}\cdot \Delta T \ \ \ /:P\\\\t = \frac{m\cdot c_{w}\cdot \Delta T}{P}[/tex]
[tex]Podstawiamy \ dane\\\\t = \frac{0,5 \ kg\cdot4200\frac{J}{kg\cdot^{o}C}\cdot80^{o}C}{1200 \ W}=\frac{168000 \ J}{1200\frac{J}{s}}\\\\\boxed{t = 140 \ s = 2 \ min \ \ 20 \ s}[/tex]
Odp. Ta grzałka doprowadzi 0,5l do wrzenia w czasie 2min 20s.
