[tex]dane:\\m = 2 \ kg\\T_1 = 20^{o}C\\T_2 = 100^{o}C\\\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 = ?\\\\\\Q = c\cdot m\cdot \Delta T\\\\Q = 4200\frac{J}{kg\cdot^{o}C}\cdot2 \ kg\cdot80^{o}C\\\\\boxed{Q = 672 \ 000 \ J = 672 \ kJ}\\\\(1 \ kJ = 1000 \ J)[/tex]
Odp. Należy dostarczyć 672 kJ ciepła.
