1.
[tex]Dane:\\m = 0,5 \ kg\\T_1 = 20^{o}C\\T_2 = 60^{o}C\\\Delta T = T_2-T_1 = 60^{o}C - 20^{o}C = 40^{o}C\\c = 180\frac{J}{kg\cdot^{o}C} \ - \ cieplo \ wlasciwe \ olowiu\\Szukane:\\Q = ?\\\\Rozwiazanie\\\\Q = m\cdot c\cdot \Delta T\\\\Q = 0,5 \ kg\cdot130\frac{J}{kg\cdot^{o}C}\cdot40^{o}C\\\\\boxed{Q = 2 \ 600 \ J = 2,6 \ kJ}\\\\(1 \ kJ = 1000 \ J)[/tex]
Odp. Potrzeba 2,6 kJ energii.
2.
[tex]Dane:\\Q = 378 \ kJ =378\cdot1000 \ J = 378 \ 000 \ J\\T_1 = 25^{o}C\\T_2 = 40^{o}C\\\Delta T = T_2 - T_1 = 40^{o}C - 25^{o}C = 15^{o}C\\c = 4200\frac{J}{kg\cdot^{o}C} \ - \ cieplo \ wlasciwe \ wody\\Szukane:\\m = ?\\\\Rozwiazanie\\\\Q = m\cdot c\cdot \Delta T \ \ \ |:(c\cdot \Delta T)\\\\m = \frac{Q}{c\cdot \Delta T}\\\\m = \frac{378000 \ J}{4200\frac{J}{kg\cdot^{o}C}\cdot15^{o}C}[/tex]
[tex]\boxed{m = 6 \ kg}[/tex]
Odp. Można podgrzać 6 kg wody.
