# Thermodynamics - Entropy

## Homework Statement

You make tea with 0.250 kg of 89.0 °C water and let it cool to room temperature (20.0 °C) before drinking it.

1)Calculate the entropy change of the water while it cools.
2)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.

## Homework Equations

Q = m*L_f (L_f = 3.34*10^5 J/kg)
Q = m*c*(delta)T

## The Attempt at a Solution

This question was posted before and this is the link

My problem is when I tried the suggested method by Andrew, what I got was
delta(S) = (0.25)(4,186)(342)/293
= 1221.5J
whereas if I used the edited solution by danni I get 244J which is given by the book.

If danni's way is correct, then could you please explain why we have to use the latent heat equation when the question doesn't involve a change in the state of the object.

Thank You

One first thing: The dimension of S is J/K, not J! Be careful, it would cost you a lot of points Danni was incorrect, Andrew was the right one. The formula was right, the only thing wrong is that you (and Danny) calculated the wrong $$\Delta T_{tea}$$.
Yup 