[tex]dane:\\m = 5 \ kg\\T_1 = 30^{o}C\\T_2 = 75^{o}C\\\Delta T = T_2 - T_1 = 75^{o}C - 30^{o}C = 45^{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}\cdot5 \ kg\cdot45^{o}C\\\\Q = 945 \ 000 \ J = 945 \ kJ[/tex]
