Change in entropy when mixing cold and hot water

[tuki]

Homework Statement

1.00 kg of hot water (T = 100 celcius) and 1.00 kg of cold water (T = 0 celcius) are mixed in bowl. It results in 2.00kg of water which temperature is 50 Celcius. Calculate the whole systems change in entropy.

Relevant Equations

Change in entropy:
$$ dS = \frac{dQ}{T} $$
Heat capacity:
$$ Q = Cm\Delta T $$

I tried following:

$$ dS_{\text{total}} = |\frac{dQ}{T_c}| |\frac{dQ}{T_H}| $$

where $T_h$ is temperature of hot water and $T_c$ is temperature of cold water. Coefficient for water wasn't provided in the assignment so i used following value c = 4190 J/kgK.

$$ dS_{\text{total}} = |\frac{cm\Delta T}{T_c}| - |\frac{cm\Delta T}{T_h}| $$

where $dQ = cm\Delta T = 4190 \text{ J/kgK} \cdot 1 \text{ kg} \cdot 50 \text{ K}$

my result for change in entropy is:

$$ dS \approx 205 \text{ J/K} $$

However the correct answer should be:

$$ dS \approx 102 \text{ J/K} $$

 

[TSny]

$$ dS_{\text{total}} = |\frac{cm\Delta T}{T_c}| - |\frac{cm\Delta T}{T_h}| $$

The expression $|\frac{cm\Delta T}{T_c}|$ is not the correct way to calculate the entropy change of the cold water. Can you see how to do this correctly by integrating the expression $dS = \frac{dQ}{T}$?

 

[tuki]

The expression $|\frac{cm\Delta T}{T_c}|$ is not the correct way to calculate the entropy change of the cold water. Can you see how to do this correctly by integrating the expression $dS = \frac{dQ}{T}$?

$$ S = \int \frac{1}{T}dQ $$

but what from now?

this equation has $\Delta T$ but not T
$$ Q = cm \Delta T $$

I think I want $T(Q)$ so I can integrate this properly?

 

[TSny]

$$ S = \int \frac{1}{T}dQ $$

Can you express $dQ$ in terms of $dT$?

 

[tuki]

Well

$$ T = \frac{dQ}{dS} \implies dT = d(\frac{dQ}{dS}) $$

Not quite sure how I would do this? (this one probably not even in right direction)

 

[TSny]

Express $\Delta Q = mc \Delta T$ in infinitesimal form.
$dQ = ?$

 

[tuki]

$dQ = mcdT$ ?

 

[TSny]

Right. So, $\int \frac{dQ}{T} = ?$

 

[tuki]

$\int \frac{dQ}{T} = \int \frac{mc}{T}dT$ ?

 

[TSny]

$\int \frac{dQ}{T} = \int \frac{mc}{T}dT$ ?

Yes. When you put appropriate limits on the integral, it will represent the change in entropy $\Delta S$ for the cold water. Deal with the hot water in a similar way.

In doing the integrals, you can treat the specific heat $c$ as a constant (independent of temperature) to a good approximation.

 

[tuki]

So then i have:

$$ \Delta S = \int_{T_0}^{T_1} \frac{cm}{T}dT = cm(ln(T_1)-ln(T_0)) $$

now when i compute $\Delta S$ for cold and hot

$$ \Delta S_{\text{total}} = cm(ln(T_n)-ln(T_H)+ln(T_c)-ln(T_n)) = cm(ln(T_c)-ln(T_h)) $$

Where $T_n$ is temperature at balanced state. Now $\Delta S_{\text{total}}$ gives wrong result

$$ \Delta S_{\text{total}} \approx -1307 \text{ J/kg} $$

Units seems to be off too.

 

[Chestermiller]

So then i have:

$$ \Delta S = \int_{T_0}^{T_1} \frac{cm}{T}dT = cm(ln(T_1)-ln(T_0)) $$

now when i compute $\Delta S$ for cold and hot

$$ \Delta S_{\text{total}} = cm(ln(T_n)-ln(T_H)+ln(T_c)-ln(T_n)) = cm(ln(T_c)-ln(T_h)) $$

Where $T_n$ is temperature at balanced state. Now $\Delta S_{\text{total}}$ gives wrong result

$$ \Delta S_{\text{total}} \approx -1307 \text{ J/kg} $$

Units seems to be off too.

The equation should read:
$$ \Delta S_{\text{total}} = cm(ln(T_n)-ln(T_H)-ln(T_c)+ln(T_n))$$
(check the signs)

For a straightforward step-by-step cookbook recipe on how to determine the change in entropy of a system that experiences and irreversible process, see the following Physics Forums Insights article: https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/

 

[TSny]

So then i have:

$$ \Delta S = \int_{T_0}^{T_1} \frac{cm}{T}dT = cm(ln(T_1)-ln(T_0)) $$

OK. $T_0$ is intial temperature and $T_1$ is final temperature.

now when i compute $\Delta S$ for cold and hot

$$ \Delta S_{\text{total}} = cm(ln(T_n)-ln(T_H)+ln(T_c)-ln(T_n)) = cm(ln(T_c)-ln(T_h)) $$

You have two sign errors in the middle expression (as pointed out by @Chestermiller).