Question 8.15: Determine the entropy produced when 3.00 g of cream at 10.0°...

First, draw a sketch of the system (Figure 8.19).

The unknown is the entropy produced by mixing, and the materials are coffee and cream, both modeled as liquid water.

Let a = cream and b = coffee. Then, \gamma = 3.00/203 = 0.0148. Assuming both the coffee and the cream are incompressible liquids with the specific heat of water, c = 4186  \text{J/(kg .K)}, Eq. (8.13) gives

_1(SP)_2= (0.203  \text{kg})[4186  \text{J/(kg .K)}]\text{ln}\left\{[ 1  +  0.0148 (\frac{10.0  +  273.15}{80.0  +  273.15} ) -1 ] × (\frac{80.0  +  273.15}{10.0  +  273.15})^{0.0148} \right\}

= 0.282  \text{J/K }

Note that this example does not include the entropy production due to the heat transfer required to cool the mixture down to a drinkable temperature.