[tex]dane:\\m = 300 \ kg\\T_1 = 20^{o}C\\T_2 = 40^{o}C\\\Delta T = T_2 - T_1 = 40^{o}C - 20^{o}C = 20^{o}C\\c = 4200\frac{J}{kg\cdot^{o}C}\\szukane:\\Q = ?\\\\Rozwiazanie\\\\Q = c\cdot m\cdot \Delta T\\\\Q = 4200\frac{J}{kg\cdot^{o}C}\cdot300 \ kg\cdot20^{o}C\\\\Q = 25 \ 200 \ 000 \ J = 25 \ 200 \ kJ = 25,2 \ MJ\\\\\\1 \ kJ = 1000 \ J\\1 \ MJ = 1 \ 000 \ 000 \ J[/tex]
Odp. Należy dostarczyć 25,2 MJ energii.
