[tex]dane:\\m = 0,5 \ kg\\T_1 = 15^{o}C\\T_2 = 45^{o}C\\\Delta T = T_2-T_1 = 45^{o}C - 15^{o}C = 30^{o}C\\t = 2 \ min =2\cdot60 \ s = 120 \ s\\P = 45 \ W\\szukane:\\c_{w} = ?\\\\Rozwiazanie\\\\P = \frac{W}{t} \ \ \rightarrow \ \ W = P\cdot t\\oraz\\W = Q = m\cdot c_{w}\cdot \Delta T\\\\m\cdot c_{w}\cdot \Delta T =P\cdot t \ \ /:(m\cdot \Delta T)\\\\c_{w} = \frac{P\cdot t}{m\cdot \Delta T}[/tex]
[tex]c_{w} = \frac{45 \ W\cdot120 \ s}{0,5 \ kg\cdot30^{o}C}=\frac{45\frac{J}{s}\cdot120 \ s}{0,5 \ kg\cdot30^{o}C}\\\\\underline{c_{w} = 360\frac{J}{kg\cdot^{o}C}}[/tex]
