[tex]Dane:\\m = 25 \ kg \ \ \ (1 \ l \ wody = 1 \ kg)\\T_1 = 20^{o}C}\\T_2 = 26^{o}C\\\Delta T = T_2-T_1 = 26^{o}C - 20^{o}C = 6^{o}C\\c = 4200\frac{J}{kg\cdot^{o}} \ - \ cieplo \ wlasciwe \ wody\\Szukane:\\Q = ?\\\\\\Q =m\cdot c\cdot \Delta T\\\\Q = 25 \ kg\cdot4200\frac{J}{kg\cdot^{o}C}\cdot6^{o}C\\\\\boxed{Q = 630 \ 000 \ J = 630 \ kJ}\\\\(1 \ kJ = 1000 \ J)[/tex]
Odp. Trzeba dostarczyć 630 kJ ciepła.
