[tex]dane:\\m = 100 \ g = 0,1 \ kg\\T_1 = 200^{o}C\\T_2 = 560^{o}\\\Delta T = T_2-T_1 = 560^{o}C - 200^{o}C = 360^{o}C\\c = 900\frac{J}{kg\cdot^{o}C} \ - cieplo \ wlasciwe \ aluminium\\szukane:\\Q = ?\\\\Q = m\cdot c\cdot \Delta T\\\\Q = 0,1 \ kg\cdot900\frac{J}{kg^{o}C}\cdot360^{o}C\\\\\underline{Q = 32 \ 400 \ J = 32,4 \ kJ}\\\\(1 \ kJ = 1000 \ J)[/tex]
Odp. Należy dostarczyć 32,4 kJ energii.
