04-06-2015, 17:26
05-06-2015, 14:31
Como entra a un recipiente rígido, no tenés entalpía saliente, solo entrante. Se trata de un recipiente adiabático, y tampoco tenés trabajo.
\[Q-W = -H_{e} + \bigtriangleup U_{s}\]
\[0 = -m_{e}*h_{e} + (U_{f}*m_{2} - U_{i}* m_{1})\]
\[(m_{2}-m_{1})*h_{e} = (U_{f}*m_{2} - U_{i}* m_{1})\]
\[(m_{2}-25kg)*510,44\tfrac{kJ}{kg} = (311,7\tfrac{kJ}{kg}*m_{2} - 211,3\tfrac{kJ}{kg}* 25kg)\]
\[m_{2}=37,63kg\]
Fin
\[Q-W = -H_{e} + \bigtriangleup U_{s}\]
\[0 = -m_{e}*h_{e} + (U_{f}*m_{2} - U_{i}* m_{1})\]
\[(m_{2}-m_{1})*h_{e} = (U_{f}*m_{2} - U_{i}* m_{1})\]
\[(m_{2}-25kg)*510,44\tfrac{kJ}{kg} = (311,7\tfrac{kJ}{kg}*m_{2} - 211,3\tfrac{kJ}{kg}* 25kg)\]
\[m_{2}=37,63kg\]
Fin