[tex]dane:\\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\\P = 530 \ W\\t = 4 \ min = 4\cdot60 \ s = 240 \ s\\c_{w} = 4200\frac{J}{kg\cdot^{o}C}\\szukane:\\m = ?\\\\Rozwiazanie\\\\W = Q = m\cdot c_{w}\cdot \Delta T\\oraz\\W = P\cdot t\\\\m\cdot c_{w}\cdot \Delta T = P\cdot t \ \ /:(c_{w}\cdot \Delta T)\\\\m = \frac{P\cdot t}{c_{w}\cdot \Delta T}[/tex]
[tex]m = \frac{530 \ W\cdot240 \ s}{4200\frac{J}{kg\cdot^{o}C}\cdot80^{o}C} = \frac{530\frac{J}{s}\cdot240 \ s}{4200\frac{J}{kg\cdot^{o}C}\cdot80^{o}C} \approx0,38 \ kg[/tex]
