[tex]Dane:\\m = 2 \ kg \ \ \ (1 \ l \ wody=1 \ kg)\\T_1 = 20^{o}\\T_2 = 100^{o}\\\Delta T = T_2-T_1 = 100^{o}C - 20^{o}C = 80^{o}C\\c = 4200\frac{J}{kg\cdot^{o}C} \ - \ cieplo \ wlasciwe \ wody\\Szukane:\\Q = ?\\\\Rozwiazanie\\\\Q = m\cdot c\cdot \Delta T\\\\Q = 2 \ kg\cdot4200\frac{J}{kg\cdot^{o}C}\cdot80^{o}C\\\\\boxed{Q = 672 \ 000 \ J = 672 \ kJ}\\\\(1 \ kJ = 1000 \ J)[/tex]
