A thermos bottle contains coffee. The thermos bottle is vigorously shaken. Consider the coffee as the system. (a) Does its temperature rise? (b) Has heat been added to it? (c) Has work been done on it? (d) Has its internal energy changed?

Jitender Pal
7 years ago
(a) Yes, its temperature rises. Since the thermos bottle is vigorously shaken, the kinetic energy of molecules in the coffee will be increased. As the temperature is directly proportional to the average kinetic energy of the molecule, therefore temperature of the coffee will be increased.
(b) No, heat has been added to the coffee. A thermos bottle contains coffee, but as the thermos bottle is insulated from its surrounding, therefore ΔQ = 0. Thus, heat has not been added to the coffee. So the sign of the work done will be negative.
(c) Yes, work has been done on the system as heat is added to the coffee. As, the thermos bottle is vigorously shaken, therefore work is done on the coffee which is against the viscous force.
(d) Yes, its internal energy has changed.Heat is the part of internal energy which is transferred from one body to another an account of the temperature difference. So the internal energy of a body was acquired by heat transfer.
In accordance to the first law of thermodynamics, ΔQ = ΔU + ΔW. From the above discussion, as, ΔQ = 0 and the sign of the work done will be negative, therefore ΔU should be positive. Thus internal energy of the coffee will be increased.
Rishi Sharma