[latex]dane:\m_1 = m_2 = m_3 = m = 2 kg\Delta T_1 = Delta T_2 = Delta T_3 = Delta T = 5^{o}C\C_1 = 380frac{J}{kg*^{o}C} - cieplo wl. miedzi\C_2 = 130frac{J}{kg*^{oC}} - cieplo wl. olowiu\C_3 = 840frac{J}{kg*^{o}C} - cieplo wl. szkla\szukane:\Q_1 = ?\Q_2 = ?\Q_3 = ?[/latex] [latex]Q = m*C*Delta T[/latex] [latex]Q_1 = m*C_1*Delta t\\Q_1 = 2kg*380frac{J}{kg*^{o}C}*5^{o}C = 3 800 J = 3,8 kJ[/latex] [latex]Q_2 = m*C_2 * Delta T\\Q_2 = 2kg*130frac{J}{kg*^{o}C}*5^{o}C = 1 300 J = 1,3 kJ[/latex] [latex]Q_3 = m*C_3 * Delta T\\Q_3 = 2kg*840frac{J}{kg*^{o}C}*5^{o} = 8 400 J = 8,4 kJ[/latex]
