[latex]Dane:\ C_{w}= 4200frac{J}{kg*^circ C} \V=0,5l=0,5dm ^{3} \ ho=1 frac{kg}{dm ^{3}} \Delta T=70^circ-20^circ=50^circ\\Szuk.\Q=?\\Roz.\Q= C_{w}*m*Delta T\\ ho= frac{m}{V} Rightarrow m= ho V\\Q= C_{w}* ho*V*Delta T\\Q=4200frac{J}{kg*^circ C} *1 frac{kg}{dm ^{3}}*0,5dm ^{3} *50^circ=105000J=105kJ[/latex]
[latex]dane:\m = 0,5 kg (dla wody 0,5l = 0,5kg)\T_1 = 20^{o}C\T_2 = 70^{o}C\Delta T = T_2 - T_1 = 50^{o}C\C = 4200frac{J}{kgcdsot ^{o}C}\szukane:\Q = ?[/latex] [latex]Q = mcdot C cdot Delta T\\Q = 0,5kgcdot4200frac{J}{kgcdot^{o}C}cdot50^{o}\\Q = 105 000 J = 105 kJ[/latex] Odp. Trzeba dostarczyć 105000 J energii (105 kJ).
