Do 5 kg wody o temperaturze 5 stoni doprowadzono 100 kg/J energi. Oblicz temperaturę końcową.

[latex]dane:\m = 5 kg\t_1 = 5^{o}C\Q = 100 kJ = 100 000 J\C = 4200 J/kg^{o}C\szukane:\t_2 = ?[/latex] [latex]Q = m * C *(t_2 - t_1)\\Q = mCt_2 - mCt_1\\mCt_2 = Q + mCt_1 /:mC\\t_2 = frac{Q + mCt_1}{mC}\\t_2 = frac{100000J + 5kg * 4200J/kgC * 5C}{5kg * 4200J/kgC} = 9,76^{o}C\\t_2 approx 9,8^{o}C[/latex]