Do 6kg wody o temperaturze 20°C wrzucono metalową kulę o masie 3kg i temperaturze 150°C. Jaka będzie temperatura końcowa wody? cw wody = 4200J/(kg∙K) cw metalu = 700J/(kg∙K

m₁ = 6 kg t₁ = 20 C cw₁ = 4200 J/kg * K = 4200 J/kg * C m₂ = 3 kg t₂ = 150 C cw₂ = 700 J/kg * K = 700 J/kg * C tk = ? Q₁ = Q₂ [latex]m_1 * c_w_1 * ( t_k - t_1 ) = m_2 * cw_2 * ( t_2 - t_k ) [/latex] [latex]6 kg * 4200 frac{J}{kg * C } * ( t_k - 20 C ) = 3 kg * 700 frac{J}{kg * C } * ( 150 C - t_k ) [/latex] [latex]25200 frac{J}{C } * ( t_k - 20 C ) = 2100 frac{J}{C } * (150 - t_k )/: 2100 frac{J}{C} [/latex] [latex] frac{25200 frac{J}{C } }{2100 frac{J}{C } } * ( t_k - 20 C ) = 150 C - t_k 12 ( t_k - 20 C ) = 150 C - t_k 12 t_k - 240 C = 150 C - t_k 12 t_k + t_k = 150 C + 240 C 13 t_k = 390 C t_k = 30 C [/latex]