Calculate the entropy change of the water while it cools

[anubis01]

Homework Statement

You make tea with 0.250 Kg of 85.0 C water and let it cool to room temperature (20.0 C) before drinking it.

a)Calculate the entropy change of the water while it cools.
b)The cooling process is essentially isothermal for the air in your kitchen. Calculate the change in entropy of the air while the tea cools, assuming that all the heat lost by the water goes into the air.
c)What is the total entropy change of the system tea + air?

Homework Equations

S=Q/T
Q=mL_f
S=mc_wln(T₂/T₁)

The Attempt at a Solution

I solved part A but I'm having trouble with part b.

a) S=mc_wln(T₂/T₁)=0.25*4190(ln(293.15/358.15))
=-210J/K

b) Since its isothermal I use the equation S=Q/T. T=20+273.15
S=Q/T=mL_f/T=(0.25*3.34x10^5)/(293.15)=285 J/K

I'm not sure if what I did for part b is correct, so if someone could tell me if I'm on the right track I would appreciate it. Thanks for the help.

 

[Andrew Mason]

Homework Equations

S=Q/T
Q=mL_f
S=mc_wln(T₂/T₁)

The Attempt at a Solution

I solved part A but I'm having trouble with part b.

a) S=mc_wln(T₂/T₁)=0.25*4190(ln(293.15/358.15))
=-210J/K

b) Since its isothermal I use the equation S=Q/T. T=20+273.15
S=Q/T=mL_f/T=(0.25*3.34x10^5)/(293.15)=285 J/K

I'm not sure if what I did for part b is correct, so if someone could tell me if I'm on the right track I would appreciate it. Thanks for the help.

Where do you get 3.34 x 10^5? You appear to be using the latent heat of fusion for water, which has no application here at all.

The heat flow, Q_air is simply -Q_water. Work that out and divide by 293K to get the change in entropy of the air.

AM