[tex]dane:\\V = 50 \ m\times21 \ m\times 1,8 \ m = 1 \ 890 \ m^{3}\\T_1 = 10^{o}C\\T_2 = 25^{o}C\\\Delta T = T_2 - T_1 = 25^{o}C - 10^{o}C = 15^{o}C\\d = 1000\frac{kg}{m^{3}} \ - \ gestosc \ wody\\c = 4200\frac{J}{kg\cdot^{o}C} \ - \ cieplo \ wlasciwe \ wody\\szukane:\\Q = ?\\\\Rozwiazanie\\\\Q = m\cdot c\cdot \Delta T\\\\d = \frac{m}{V} \ \ \rightarrow \ \ m = V\cdot d = 1890 \ m^{3}\cdot1000\frac{kg}{m^{3}} = 1 \ 890 \ 000 \ kg\\\\Q = 1890000 \ kg\cdot4200\frac{J}{kg\cdot^{o}C}\cdot15^{o}C[/tex]
[tex]Q = 119 \ 070 \ 000 \ 000 \ J = 119 \ 070 \ 000 \ kJ = 119 \ 070 \ MJ = 119,07 \ GJ\\\\\\1 \ kJ = 1000 \ J\\1 \ MJ = 1 \ 000 \ 000 \ J\\1 \ GJ = 1 \ 000 \ 000 \ 000 \ J[/tex]
