Question 2.10: Cooling of a Hot Fluid in a Tank A rigid tank contains a hot...

A fluid in a rigid tank looses heat while being stirred. The final internal energy of the fluid is to be determined.
Assumptions 1 The tank is stationary and thus the kinetic and potential energy changes are zero, \Delta KE =\Delta PE =0 . \text { Therefore, } \Delta E=\Delta U and internal energy is the only form of the system’s energy that may change during this process. 2 Energy stored in the paddle wheel is negligible.
Analysis Take the contents of the tank as the system (Fig. 2–47). This is a closed system since no mass crosses the boundary during the process. We observe that the volume of a rigid tank is constant, and thus there is no moving boundary work. Also, heat is lost from the system and shaft work is done on the system. Applying the energy balance on the system gives

\underbrace{E_{\text {in }}-E_{\text {out }}}_{\begin{array}{c}\text { Net energy transfer } \\\text { by heat, work, and mass }\end{array}}=\underbrace{\Delta E_{\text {system }}}_{\begin{array}{c}\text { Change in internal, kinetic, } \\\text { potential, etc., energies }\end{array}}

\begin{aligned}W_{\text {sh,in }}-Q_{\text {out }} &=\Delta U=U_{2}-U_{1} \\100 kJ -500 kJ &=U_{2}-800 kJ \\U_{2} &=400 kJ\end{aligned}

Therefore, the final internal energy of the system is 400 kJ.