[tex]Dane:\\m = 1 \ kg\\T_1 = 6^{o}C\\T_2 = 100^{o}C\\\Delta T = T_2-T_1 = 100^{o}C - 6^{o}C = 94^{o}C\\c = 4200\frac{J}{kg\cdot^{o}C} \ - \ cieplo \ wlasciwe \ wody\\Szukane:\\Q = ?\\\\Rozwiazanie\\\\Q = m\cdot c\cdot \Delta T\\\\Q = 1 \ kg\cdot4200\frac{J}{kg\cdot^{o}C}\cdot94^{o}\\\\\boxed{Q = 394 \ 800 \ J = 394,8 \ kJ}[/tex]
[tex](1 \ kJ = 1000 \ J)[/tex]
Odp. Trzeba dostarczyć 394,8 kJ energii.
