Oblicz temperaturę poczatkową wody o masie 1 kg , jeżeli po dostarczeniu jej 300 kJ ciepła jej temperatura wzrosła o 95°C .

m = 1 kg Q = 300 kJ = 300000 J      bo 1 kJ = 1000 J cw = 4200 J/kg * C tk = 95 C tp = ? [latex]Q = m * c_w * ( t_k - t_p) /: m * c_w[/latex] [latex]t_k - t_p = frac{Q }{m * c_w} [/latex] [latex]95 C - t_p = frac{300000 J }{1 kg * 4200 frac{J}{kg * C } } [/latex] [latex]95 C - t_p = 71,43[/latex] [latex]t_p = 95 C - 71,43 C = 23,57 C [/latex]

[latex]dane:\m = 1 kg\Q = 300 kJ = 300 000 J\Delta T =T_2 - T_1 = 95^{o}C\C = 4200frac{J}{kgcdot^{o}C}\szukane:\T_1 = ?[/latex] [latex]Q = mcdot C cdot(T_2-T_1) |:mC\\ frac{Q}{mcdot C}=T_2-T_1\\T_1 = T_2 - frac{Q}{mC}\\T_1= 95^{o}-frac{300000J}{1kgcdot 4200frac{J}{kgcdot^{o}C}}\\T_1 = 95^{o}C-71.4^{o}C\\T_1 approx23,6^{o}C[/latex]