[latex]m = 0,5 kg T_1 = 0 C T_2 = 20 C Delta T = 20 C - 0 C = 20 C c_t = 334000 frac{J}{kg} c_w = 4200 frac{J}{kg * C } Q = ? Q= Q_1 + Q_2 Q_1 = m * c_t Q_1 = 0,5 kg * 334000 frac{J}{kg} Q_1 = 167000 j = 167 kJ[/latex] [latex]Q_2 = m * c_w *Delta T Q_2 = 0,5 kg * 4200 frac{J}{kg * C } *20 C Q_2 = 42000 J = 42 kJ Q = 209 kJ[/latex] zakładam, że temp. pokojowa wynosi 20 C
[latex]dane:\m = 0,5 kg\t_1 = 0^{o}C\t_2 = 20^{o}C\Delta T = 20^{o}C\C_{t} = 334000 J/kg\C_{w} = 4200 J/kg^{o}C\szukane:\Q = ?[/latex] [latex]Q = Q_{1} + Q_2[/latex] Q1 - ciepło potrzebne do stopienia lodu o masie 0,kg [latex]Q_1 = m * C_{t} = 0,5kg * 334000J/kg\Q1 = 167000 J = 167 kJ[/latex] Q2 - ciepło potrzebne do podgrzania stopionego lodu (wody) do temp. 20*C [latex]Q_2 = m * C_{w} * Delta T = 0,5kg * 4200J/kg^{o}C * 20^{o}\Q2 = 42000 J = 42 kJ[/latex] [latex]Q = 167kJ + 42kJ\\Q = 209 kJ[/latex]
