Why does entropy increase when hot water is mixed with cold?

[Engineer1]

Why is entropy lost by hot water less than the entropy gained by the cold water?From perspective of energy,why is it better to take water and heat it to a temperature than it is to mix hot water and cold water to get a particular temperature.

 

[jedishrfu]

Welcome to PF!

In weather they often say cold air masses don't mix well with warm air masses as the densities are different. In general, cold air masses push warm ones out of an area.

So I imagine the same goes for liquids when you mix different temperatures then you'll have an uneven distribution of heat whereas with heating and convection the distribution will be more uniform over time.

 

[Engineer1]

Thank you.But mixing of waters of same mass but different temperature gives water with resultant temperature which is average of those temperatures.So,there is equal temperature in the resultant mixed water.It is still confusing.

 

[jedishrfu]

But that's over time, initially and for some short time thereafter there are different temps in the combined liquid. The temp is not uniform throughout.

 

[Engineer1]

But heating of water also takes time in reaching a particular temperature.

 

[Engineer1]

Also,why does 1kJ heat from 1000 kelvin reservoir has less entropy than 1kJ heat from 300 kelvin reservoir?Less entropy means more orderliness and more information in the system.

 

[DrClaude]

Why is entropy lost by hot water less than the entropy gained by the cold water?

Also,why does 1kJ heat from 1000 kelvin reservoir has less entropy than 1kJ heat from 300 kelvin reservoir?Less entropy means more orderliness and more information in the system.

The "entropy as disorder" may be ok for popular science, but it won't get you far. It is better to see it in terms of multiplicity, using Boltzmann's famous equation: $S = k \ln \Omega$. The multiplicity $\omega$ is the number of microstates corresponding to a given macrostate, or in simpler (but imperfect) language: how many ways can the energy be distributed among all the degrees of freedom of the system.

When you remove a given amount of energy from a system that has a lot of it (high T), you are reducing the number of ways the energy that is left can be organised, but not as much as the increase in number of ways that lower energy system (low T) can now reorganise all its energy. This is a simple way of representing why $\Delta S = q/T$.

From perspective of energy,why is it better to take water and heat it to a temperature than it is to mix hot water and cold water to get a particular temperature.

From the point of view of energy, if the final quantity and temperature of the water is the same, I don't see any difference.

 

[Engineer1]

Thank you.

 

[Engineer1]

Here is some more stuff which I found relevant

 

[Dale]

Why is entropy lost by hot water less than the entropy gained by the cold water?

Remember that $dS=dQ/T$. So if the hot object loses energy dQ then that is a small loss of entropy since its T is large, meanwhile the cold object gains dQ and therefore gains a lot of entropy since its T is small.

 

[Chestermiller]

@Engineer1 It seems (at least to me) that you are not being clear about the comparison you are trying to make. You seem to be looking at two different situations, and trying to compare them. For each situation, please specify the initial an final thermodynamic equilibrium states of the systems. For example:

COMPARE THE ENTROPY CHANGES FOR SITUATIONS 1 AND 2
Situation 1
State 1: mass m of liquid at temperature $T_{hot}$, and mass m of liquid at temperature $T_{cold}$
State 2: mass 2m of liquid at temperature $(T_{hot}+T_{cold})/2$

Situation 2
State 1: ?
State 2: ?

 

[Engineer1]

@Engineer1 It seems (at least to me) that you are not being clear about the comparison you are trying to make. You seem to be looking at two different situations, and trying to compare them. For each situation, please specify the initial an final thermodynamic equilibrium states of the systems. For example:

COMPARE THE ENTROPY CHANGES FOR SITUATIONS 1 AND 2
Situation 1
State 1: mass m of liquid at temperature $T_{hot}$, and mass m of liquid at temperature $T_{cold}$
State 2: mass 2m of liquid at temperature $(T_{hot}+T_{cold})/2$

Situation 2
State 1: ?
State 2: ?

COMPARE THE ENTROPY CHANGES FOR SITUATIONS 1 AND 2
Situation 1
State 1: mass m 1kg of liquid at temperature $T_{hot}$ 60 degrees centigrade, and mass m 1kg of liquid at temperature $T_{cold}$ 10 degrees centigrade
State 2: mass 2m of liquid at temperature $(T_{hot}+T_{cold})/2$

 

[Chestermiller]

COMPARE THE ENTROPY CHANGES FOR SITUATIONS 1 AND 2
Situation 1
State 1: mass m 1kg of liquid at temperature $T_{hot}$ 60 degrees centigrade, and mass m 1kg of liquid at temperature $T_{cold}$ 10 degrees centigrade
State 2: mass 2m of liquid at temperature $(T_{hot}+T_{cold})/2$

So, do you know how to calculate the entropy change for this situation?