[tex]dane:\\\eta = 80\% = 0,8\\V = 25 \ m\times 10 \ m\times 2 m = 500 \ m^{3}\\t = 40 \ h = 40\cdot3600 \ s =144 \ 000 \ s\\T_1 = 5^{o}C\\T_2 = 45^{o}C\\\Delta T = T_2 - T_1 = 45^{o}C - 5^{o}C = 40^{o}C\\d = 1000\frac{kg}{m^{3}} \ - \ gestosc \ wody\\c = 4200\frac{J}{kg\cdit^{o}C} \ - \ cieplo \ wlasciwe \ wody\\szukane:\\P = ?\\\\Rozwiazanie\\\\P = \frac{W}{t}\\\\W = Q = m\cdot c \cdot \Delta T\\\\m = V\cdot d\\\\W = V\cdot d\cdot c\cdot \Delta T[/tex]
[tex]W = 50 \ m^{3}\cdot1000\frac{kg}{m^{3}}\cdot4200\frac{J}{kg\cdot^{o}C}\cdot 40^{o}C}=8 \ 400 \ 000 \ 000 \ J\\\\P = \frac{\eta\cdot W}{t}\\\\P = \frac{08\cdot8 \ 400 \ 000 \ 000 \ W}{144 \ 000 \ s} = 46 \ 666,(6) \ W \approx 46,7 \ kW[/tex]
Odp. Moc nagrzewnicy P ≈ 46,7 kW.