A liquid is irregularly stirred in a well-insulated container and thereby undergoes a rise in temperature. Regard the liquid as the system. Has heat been transferred? How can you tell? Has work been done? How can you tell? Why is it important that the stirring is irregular? What is the sign of ?U? How can you tell?
Solution to 13DQ Step 1 In the question it has been mentioned that the container is well insulated, thus the volume of the container doesn’t change. Thus the process of stirring the container filled with water is an isochoric process and an adiabatic transformation of work. The process is irreversible as well because stirring is a work done on viscous forces, thus it is a non reversible process. (a) No, the heat has not been transferred as the container is a well insulated one therefore heat cannot enter nor it can leave the container, thus dQ=0. (b) Yes, the work has been done on the system. Stirring will put the system out of equilibrium. Stirring is a work done against viscous forces of the liquid, thus work has been done on the system. (c) Stirring is irregular in order to maintain the final and initial states of the system in equilibrium. Stirring is a irreversible process. Thus, in order to find the entropy change in the system integration cannot be performed. Thus we will assume a path R which will take the system from initial equilibrium state to final equilibrium state. R is any arbitrarily chosen reversible process. (d) Entropy of a irregularly stirred, thermally insulated system of fluid is given by: T2 dQ =S T1 T dQ=C p S=C pT2T1 Thus entropy is a positive. U=dQ-dW dQ=0, Thus, U is negative