Do naczynia wlano 50g nawty. Dostaeczono 105J energii. Nawta ogrzała sie o 1*C. Oblicz ciepło właściwe nawty.

[latex]m = 50 g = 0,05 kg Q = 105 J Delta T = 1 C c_w = ? Q = m * c_w *Delta T c_w = frac{Q}{m*Delta T } c_w = 105 J /0,05 kg *1 C c_w = 2100 frac{J}{kg * C } [/latex]

[latex]dane:\m = 50 g= 0,05 kg\Q = 105 J\Delta T = 1^{o}C\szukane:\C_{w} = ?[/latex] [latex]Q = C_{w} * m * Delta T /:mDelta T\\C_{w} = frac{Q}{m * Delta T} = frac{105J}{0,05kg * 1^{o}C}}\\C_{w} = 2100 J/kg*^{o}C[/latex]