[tex]dane:\\Q = 10 \ kJ = 10 \ 000 \ J\\T_1 = 15^{o}C\\T_2 = 35^{o}C\\\Delta T = T_2 - T_1 = 35^{o}C - 15^{o}C = 20^{o}C\\c = 4200\frac{J}{kg\cdot^{o}C} \ - \ cieplo \ wlasciwe \ wody\\szukane:\\m = ?\\\\Rozwiazanie\\\\Q = c\cdot m\cdot \Delta T \ \ /:(c\cdot \Delta T)\\\\m = \frac{Q}{c\cdot \Delta T}\\\\m = \frac{10000 \ J}{4200\frac{J}{kg\cdot^{o}C}\cdot20^{o}C0,119 \ kg}\\\\m \approx 0,119 \ kg[/tex]
Odp. Można podgrzać ok. 0,119 kg wody.
