Do 5 litrow wody o temperaturze 20°C wlano 1 kg wody o temperaturze 100°C. Oblicz temp. koncowa.

[latex]V_{1}=0,005 [m^{3}]\\d=1000 [frac{kg}{m^{3}}]\\m_{2}=1 [kg]\\T_{1}=20 [^{o}C]\\T_{2}=100 [^{o}C][/latex] Masa wody: [latex]m_{1}=dV_{1}=5 [kg][/latex] Układamy bilans cieplny i wyznaczamy temperaturę końcową wody: [latex]Q_{1}+Q_{2}=0\\m_{1}c_{w}(T_{k}-T_{1})+m_{2}c_{w}(T_{k}-T_{2})=0\\m_{1}(T_{k}-T_{1})+m_{2}(T_{k}-T_{2})=0\\T_{k}=frac{m_{1}T_{1}+m_{2}T_{2}}{m_{1}+m_{2}}=33frac{1}{3} [^{o}C][/latex]

[latex]dane:\m_1 = 5 kg (1 l wody = 1 kg)\t_1 = 20^{o}C\m_2 = 1 kg\t_2 = 100^{o}C\szukane:\t_{k} = ?[/latex] [latex]BILANS ENERGII WEWNETRZNEJ\\Q_{pobrane} = Q_{oddane}\\m_1C(t_{k}-t_1) = m_2C(t_2-t_{k}) /:C\\m_1(t_{k}-t_1) = m_2(t_2-t_k})\\m_1t_{k}-m_1t_1=m_2t_2-m_2t_{k}\\m_1t_{k}+m_2t_{k} = m_1t_1+m_2t_2\\t_{k}(m_1+m_2) = m_1t_1+m_2t_2 /:(m_1+m_2)[/latex] [latex]t_{k} = frac{m_1t_1+m_2t_2}{m_1+m_2} = frac{5kg*20^{o}C+1kg*100^{o}C}{5kg+1kg} = frac{200}{6} ^{o}C\\t_{k} = 33,(3) ^{o}Capprox33,3 ^{o}C[/latex]