[tex]dane:\\m = 1 \ kg\\Q = 220 \ kJ = 220 \ 000 \ J\\c = 4200\frac{J}{kg\cdot K}=4200\frac{J}{kg\cdot^{o}C}\\szukane:\\\Delta T=?\\\\Rozwiazanie\\\\Q = c\cdot m\cdot \Delta T \ \ /:(c\cdot m)\\\\\Delta T = \frac{Q}{c\cdot m}\\\\\Delta T = \frac{220000 \ J}{4200\frac{J}{kg\cdot^{o}C}\cdot1 \ kg}\\\\\Delta T \approx 52,4^{o}C[/tex]
Odp. Temperatura wody wzrośnie o ok. 52,4°C.
