[tex]dane:\\m = 8 \ kg \ \ (1 \ l \ wody = 1 \ 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}\\szukane:\\Q = ?\\\\Q = c\cdot m\cdot \Delta T\\\\Q = 4200\frac{J}{kg\cdot^{o}C}\cdot8 \ kg\cdot80^{o}C\\\\Q = 2 \ 688 \ 000 \ J = 2 \ 688 \ kJ = 2,688 \ MJ\\\\1 \ kJ = 1000 \ J\\1 \ MJ = 1 \ 000 \ 000 \ J[/tex]
Odp. Woda pobierze 2,688 MJ energii.
