You have two styrofoam containers of water. Each holds 1 kg of water. In one the water temperature is 17°C, while in the other it is 37°C. The colder water is then poured into the warmer water, and the system is allowed to come to equilibrium.

Is this process reversible or irreversible?

1. Reversible (5/33) (15%)
2. Irreversible (28/32) (85%)

The process is irreversible - any process involving a transfer of heat from a higher-temperature region to a lower-temperature region is irreversible.

Assuming no heat is exchanged with the surroundings or the environment, what is the change in entropy in the mixing process?

First, determine how much heat is involved. The final temperature will be 27°C. The heat transferred from the hot water to the cold water is therefore:

m = 1, c = 4186, and ΔT = 10°.

Q = mcΔT = 41860 J

Calculate the change in entropy for the hot and cold water using the equation:

ΔS

=

Q Tavg

The T's are in kelvin. For the cold water, T_avg = 22°C = 295 K. For the hot water T_avg = 32°C = 305 K.

Q is positive for the cold water, because heat was added, and negative for the hot water. Therefore, for the mixing process:

ΔScold + ΔShot

=

41860 295

-

41860 305

= +4.65 J/K