[tex]dane:\\V = 25 \ m \times 10 \ m\times 2 \ m = 500 \ m^{3} = 500 \ 000 \ dm^{3} = 500 \ 000 \ l\\m = 500 \ 000 \ kg \ \ (1 \ l \ wody = 1 \ kg)\\T_1 = 15^{o}C\\T_2 = 30^{0}C\\\Delta T = T_2 - T_1 = 30^{o}C - 15^{o}C = 15^{o}C\\c = 4200\frac{J}{kg\cdot^{o}C} \ - \ cieplo \ wlasciwe \ wody\\1 \ kWh = 3 \ 600 \ 000 \ J\\k = 0,50 \ zl/1kWh\\szukane:\\K = ?[/tex]
Rozwiązanie
[tex]Q = c\cdot m\cdot \Delta T\\\\Q = 4200\frac{J}{kg\cdot^{o}C}\cdot500000 \ kg\cdot15^{o}C = 31 \ 500 \ 000 \ 000 \ J\\\\Q = \frac{31500000000 \ J}{3600000\frac{J}{kWh}} = 8 \ 750 \ kWh\\\\\\K = k\cdot Q\\\\K = 0,50 \ zl/kWh\cdot8750 \ kWh = 4 \ 375 \ zl[/tex]
Odo. Koszt podgrzewania tej wody, to 4 375 zł.
