[tex]dane:\\m = 2 \ kg\\T_1 = 20^{o}C\\T_2 = 1020^{o}C\\\Delta T = T_2 - T_1 \\c_{w} =460\frac{J}{kg\cdot^{o}C} \ - cieplo \ wlasciwe \ stali\\szukane:\\Q = ?\\\\Q = m\cdot c_{w} \cdot (T_2-T_1)\\\\Q = 2 \ kg\cdot460\frac{J}{kg\cdot^{o}C}\cdot(1020^{o}C - 20^{o}C)\\\\\underline{Q = 920 \ 000 \ J = 920 \ kJ}\\\\(1 \ kJ = 1000 \ J)[/tex]
Odp. Potrzebna jest ilość ciepła Q = 920 kJ.
