Calculating Entropy Change in a Water Mixing Experiment

[Vuldoraq]

Homework Statement

In order to take a nice warm bath, you mix 60 litres of hot water at 55◦C with 30 litres of
cold water at 10◦C. How much new entropy have you created by mixing the water?

Homework Equations

Q=mc\Delta{T}
dS=dQ/T

The Attempt at a Solution

Greetings!

I am completely stuck on this question. I just can't work how I am supposed to find the change in internal energy of the system. I am geussing that T should be the final temperature of the whole system, but how to find it I do not know. If someone could please point me in the right direction I would very grateful!

Thanks

 

[Andrew Mason]

Homework Statement

In order to take a nice warm bath, you mix 60 litres of hot water at 55◦C with 30 litres of
cold water at 10◦C. How much new entropy have you created by mixing the water?

Homework Equations

Q=mc\Delta{T}
dS=dQ/T

The Attempt at a Solution

Greetings!

I am completely stuck on this question. I just can't work how I am supposed to find the change in internal energy of the system. I am geussing that T should be the final temperature of the whole system, but how to find it I do not know. If someone could please point me in the right direction I would very grateful!

Thanks

It does not ask for the change in internal energy. (That is complicated for a liquid). It asks for the total change in entropy. What you have to do is work out the heat flow divided by temperature from the hot water to the cold water from the initial to final states. To do this, you have to determine the final temp. Then you have to do two integrations to find the entropy change (you integrate dQ/T over the reversible path from the unmixed water to the mixed water).

\Delta S = \Delta S_{hot} + \Delta S_{cold}

To help you, I will give you the integral for the change in entropy of the hot water:

\Delta S_{hot} = \int_{60}^{T_f} dQ/T

You just have to express dQ in terms of heat capacity and dT and do the integrations.

AM

 

[Vuldoraq]

Hi Andrew, thanks for your speedy reply,

Now, dQ=cdT so the integral becomes,

\int_{60}^{30}c/T dT=-cln2

I am confused about the limits though, since they appear to be the volumes of the water? Has the latex gone wrong?

 

[Vuldoraq]

The latex seems to have corrected itself now.

So, redoing the above,

\int_{55}^{T_{f}}c/TdT=c(lnT_{f}-ln55)
\int_{10}^{T_{f}}c/TdT=c(lnT_{f}-ln10

Adding gives the total Entropy,

S=c(2lnT_{f}-ln550)

How do I find the final temperature?

 

[Andrew Mason]

The latex seems to have corrected itself now.

So, redoing the above,

\int_{55}^{T_{f}}c/TdT=c(lnT_{f}-ln55)
\int_{10}^{T_{f}}c/TdT=c(lnT_{f}-ln10

Adding gives the total Entropy,

S=c(2lnT_{f}-ln550)

How do I find the final temperature?

That is the easiest part. You know there is no net heat flow so the heat flow out of the warmer water has to equal the heat flow into the cooler water. They have the same heat capacity, so:

m_{warm}C\Delta T_{warm} = - m_{cool}C\Delta T_{cool}

and: \Delta T_{cool} - \Delta T_{warm} = (T_f - 10) - (T_f - 55) = 45

AM

 

[Vuldoraq]

Excellent. Thankyou very much for your help!

Vuldoraq.