[tex]dane:\\m_1 = 0,5 \ kg\\T_1 = 0^{o}C\\T_2 = 40^{o}C\\\Delta T = T_2 - T_1 = 40^{o}C - 0^{o}C = 40^{o}C\\T_{k} = 30^{o}C\\szukane:\\m_2 = ?\\\\Rozwiazanie\\\\Q_{pobrane} = Q_{oddane}\\\\m_1C(T_{k}-T_1) = m_2C(T_{2}-T_{k}) \ /:C(T_2 - T_{k})\\\\m_2 = m_1\cdot\frac{T_{k}-T_1}{T_2-T_{k}}[/tex]
[tex]m_2 = 0,5 \ kg \cdot\frac{30^{o}C - 0^{o}C}{40^{o}C - 30^{o}C}\\\\m_2 = 0,5 \ kg\cdot 3\\\\\boxed{m_2 = 1,5 \ kg}[/tex]
Odp. Masa dolanej wody m₂ = 1,5 kg.
