[Gold Sky]

Hello,

As what we know, there are two types of the measurable properties:
1- Extensive like mass, volume and length. 
2- Intensive like Density and Temperature. 

Extensive properties can be added. For example, you have 1 kg of water. You add more 1 kg of water. Thus, you have 2 kg of water. On the other hand, density and temperature cannot be simply added. For example, if you have water with X degrees temperature, and you have the same amount of water with X degrees temperature, too. Adding them will not make the temperature 2X degrees. 

But... (here my question ) 

what If I have, for example, 5 C° as first water and 6 C° as second water. what will be the temperature if I am gonna add them ?

My personal answer: I think it will be the largest available temperature  which is 6 C°
Is my understanding correct ? 

The same thing for density .
what will happen if I am gonna add 2 diff. densities of liquids, for example ? Will the final density will be the largest density or what ?
The reason of this question: I am solving a problem depends on this little fact which Idk it 


Thanks,
Gold Sky

[sjb]

Hello,

As what we know, there are two types of the measurable properties:
1- Extensive like mass, volume and length. 
2- Intensive like Density and Temperature. 

Extensive properties can be added. For example, you have 1 kg of water. You add more 1 kg of water. Thus, you have 2 kg of water. On the other hand, density and temperature cannot be simply added. For example, if you have water with X degrees temperature, and you have the same amount of water with X degrees temperature, too. Adding them will not make the temperature 2X degrees. 

But... (here my question ) 

what If I have, for example, 5 C° as first water and 6 C° as second water. what will be the temperature if I am gonna add them ?

My personal answer: I think it will be the largest available temperature  which is 6 C°
Is my understanding correct ? 

The same thing for density .
what will happen if I am gonna add 2 diff. densities of liquids, for example ? Will the final density will be the largest density or what ?
The reason of this question: I am solving a problem depends on this little fact which Idk it 


Thanks,
Gold Sky

For part 1, you can think of the liquids having a heat content, so assuming you have the same liquid (for reasons that will become apparent), if you have 1 litre @ 5°C and 3 litre @ 6°C, you have a total of 2 litres at (5+6)/(1+3) °C, so what temperature?

For part 2, look into ideal and non-ideal solutions. 1 litre of water and 1 litre of ethanol do not make 2 litres of mix, even if the masses are additive.

[Gold Sky]

Hello,

As what we know, there are two types of the measurable properties:
1- Extensive like mass, volume and length. 
2- Intensive like Density and Temperature. 

Extensive properties can be added. For example, you have 1 kg of water. You add more 1 kg of water. Thus, you have 2 kg of water. On the other hand, density and temperature cannot be simply added. For example, if you have water with X degrees temperature, and you have the same amount of water with X degrees temperature, too. Adding them will not make the temperature 2X degrees. 

But... (here my question ) 

what If I have, for example, 5 C° as first water and 6 C° as second water. what will be the temperature if I am gonna add them ?

My personal answer: I think it will be the largest available temperature  which is 6 C°
Is my understanding correct ? 

The same thing for density .
what will happen if I am gonna add 2 diff. densities of liquids, for example ? Will the final density will be the largest density or what ?
The reason of this question: I am solving a problem depends on this little fact which Idk it 


Thanks,
Gold Sky

For part 1, you can think of the liquids having a heat content, so assuming you have the same liquid (for reasons that will become apparent), if you have 1 litre @ 5°C and 3 litre @ 6°C, you have a total of 2 litres at (5+6)/(1+3) °C, so what temperature?

For part 2, look into ideal and non-ideal solutions. 1 litre of water and 1 litre of ethanol do not make 2 litres of mix, even if the masses are additive.

First of all, thank you for answering my questions and I really appreciate this.

For part 1,
to be honest, I did not get it because the unit of (5+6)/(1+3) is L/C°, not C°. But let me revise my question: the same question but assume the volumes are the same !!!! what is the result ?

For part 2,
I got it but..
2.4- I did not understand the expression "ideal and non-ideal". DO you mean homo and heterogeneous? 
2.5-  what will be the final density ? I am really stuck with this point for 3 hours 
2.6- I have two diff. liquids with two diff. densities. Let's say A and B.
I added them and they performed a homogeneous solution. The volume of the solution is 60 percent volume , for example, by A and 40 percent volume by B. Thus, assume the total volume is 100 L. Thus, A has 60 L and B has L. The total is 100 L. So, what can you say now if A is water and b is methanol ? 

Thanks,

[Irlanur]

what If I have, for example, 5 C° as first water and 6 C° as second water. what will be the temperature if I am gonna add them ?

My personal answer: I think it will be the largest available temperature  which is 6 C°
Is my understanding correct ? 

As very often in chemsitry, what does your experience tell you? think of 1 L boiling water and 1 L with nearly 0 celcius. do you really think that you will have 2 L of boiling water? that would be great because you could heat up one drop of water and then just add drop after drop till you have enough boiling water for spaghetti without heating anymore. But unfortunately, that's not the case.

[Gold Sky]

"As very often in chemsitry, what does your experience tell you? think of 1 L boiling water and 1 L with nearly 0 celcius. do you really think that you will have 2 L of boiling water? that would be great because you could heat up one drop of water and then just add drop after drop till you have enough boiling water for spaghetti without heating anymore. But unfortunately, that's not the case."

Got it 

what about the density ?

[sjb]

For part 1, you can think of the liquids having a heat content, so assuming you have the same liquid (for reasons that will become apparent), if you have 1 litre @ 5°C and 3 litre @ 6°C, you have a total of 2 litres at (5+6)/(1+3) °C, so what temperature?

For part 2, look into ideal and non-ideal solutions. 1 litre of water and 1 litre of ethanol do not make 2 litres of mix, even if the masses are additive.

First of all, thank you for answering my questions and I really appreciate this.

For part 1,
to be honest, I did not get it because the unit of (5+6)/(1+3) is L/C°, not C°. But let me revise my question: the same question but assume the volumes are the same !!!! what is the result ?

For part 2,
I got it but..
2.4- I did not understand the expression "ideal and non-ideal". DO you mean homo and heterogeneous? 
2.5-  what will be the final density ? I am really stuck with this point for 3 hours 
2.6- I have two diff. liquids with two diff. densities. Let's say A and B.
I added them and they performed a homogeneous solution. The volume of the solution is 60 percent volume , for example, by A and 40 percent volume by B. Thus, assume the total volume is 100 L. Thus, A has 60 L and B has L. The total is 100 L. So, what can you say now if A is water and b is methanol ? 

Thanks,

Sorry, typo'd slightly in my explanation. I was trying to introduce the idea of a weighted average.

As to part 2, no, I'm not sure I meant either hetero- or homogeneous solutions. But I think if you do have a homogeneous solution as you describe then you can use weighted averages again.