[tex]dane:\\m_1 = 840 \ kg\\h = 10 \ m\\T_1 = 20^{o}C\\T_2 = 100^{o}C\\\Delta T = T_2-T_1 = 100^{o}C - 20^{o}C = 80^{o}C\\c = 4200\frac{J}{kg\cdot ^{o}C} - \ cieplo \ wlasciwe \ wody\\g = 10\frac{N}{kg}\\szukane:\\m_2= ?\\\\Rozwiazanie\\\\E_{p} = m_1\cdot g\cdot h\\\\Q = c\cdot m_2\cdot \Delta T\\\\Q = E_{p}\\\\c\cdot m_2 \cdot \Delta T=m_1\cdot g\cdot \ h \ \ /:(c\cdot \Delta T)\\\\m_2 = \frac{m_1\cdot \ g\cdot \ h}{ c\cdot \Delta T}[/tex]
[tex]m_2 = \frac{840 \ kg\cdot10\frac{N}{kg}\cdot 10 \ m}{4200\frac{J}{kg\cdot ^{o}C}\cdot80^{o}} = 0,25 \ kg[/tex]
Odp. Można zagotować 0,25 kg (0,25 l) wody.
