Answer to Question #90494 in Molecular Physics | Thermodynamics for an

Question

Making a glass of Brown Sugar Bubble Milk uses 1/4 L of 85.0°C water before cooling it down to room temperature at 27.0°C.
a) Calculate the entropy change of the milk while it cools, the specific heat of water is 4190 J·g-1·K-1
b) Calculate the change in entropy of the air while the milk cools, if all the heat lost by the water goes into the air, and the cooling process is virtually isothermal for the air.
c) What is the total entropy change of the system (milk + air)? Explain the meaning of the result

Answer 1

a) The entropy change is equal to

$$ΔS= \int_ { T_1 }^ {T_2} \frac {dQ} {T} (1)$$

The heat change is given by formula

$$dQ=cmdT= cVρdT (2)$$

We put (2) in (1):

$$ΔS= \int_ { T_1 }^ {T_2} \frac { cVρdT } {T} (3)$$

In our case, we have

T₁=358 K, T₂=300 K, V=0.25×10^-3 m³, c=4190 J/kg×K

Using (3) we get:

ΔS=-185 J/K

Answer:

-185 J/K

b) The entropy change is equal to

$$ΔS= \frac { cVρ ΔT } {T_2} (1)$$

where ΔT=358 K ­300 K=58 K, T₂=300 K

Using (3) we get:

ΔS=203 J/K

Answer:

203 J/K

c)

ΔS=-185 J/K+203 J/K = 18 J/K

Answer:

18 J/K

The total entropy change is positive, as expected (irreversible process).