[tex]dane:\\T_1 = 20^{o}C\\m = 2 \ kg\\Q = 20 \ kJ = 20 \ 000 \ J\\c = 4200\frac{J}{kg\cdot^{o}C} \ - \ cieplo \ wlasciwe \ wody\\szukane:\\T_2=?\\\\Rozwiazanie\\\\Q = c\cdot m\cdot (T_2-T_1) \ \ /:(c\cdot m)\\\\T_2 - T_1 = \frac{Q}{c\cdot m}\\\\T_2 = T_1 + \frac{Q}{c\cdot m}\\\\T_2 = 20^{o}C + \frac{20 \000 \ J}{4200\frac{J}{kg\cdot^{o}C}\cdot2 \ kg}\\\\T_2 = 20^{o}C + 2,4^{o}C\\\\\boxed{T_2 = 22,4^{o}C}[/tex]
Odp. Woda ogrzeje się do temperatury T₂ = 22,4°C.
