What is the final temperature when mixing heated and cool water?

[chem_tr]

I haven't found my textbooks on physical chemistry, so I want to ask the forum members.

When we heat 4 liters of water from 10°C to 90°C, and add this to 16 liters of water at 10°C, what will the final temperature be?

My brainstorming revealed the following, but I'm not sure if they are correct or not (my physical chemistry really sucks):

If we heat 4 liters of water to 90°C, we must give an energy which equals Q=m \times c \times \Delta t=4000 \times 1 \times (90-10)=320000~cal. This amount of energy is dissipated in the whole pool of water, so 320000=16000 \times 1 \times (t_{final}-10)=>t_{final} is found to be 30°C

I am almost sure it is wrong, so could anybody please show me the very logic of this type of calorimetric problem?

Thank you.

 

[Bystander]

You have a total of 20 liters, not 16. That help?

 

[GCT]

When we heat 4 liters of water from 10°C to 90°C, and add this to 16 liters of water at 10°C, what will the final temperature be?

mc(x-90)=m_2c(x-10)

-x represents the equilibrium temperature

-heat lost by the 90 degree water is gained by 10 degree water

I'm pretty sure this is correct

 

[Sirus]

Let me try:

(\mbox{heat lost})=-(\mbox{heat gained})
(mc\Delta t)_{1}=-(mc\Delta t)_{2}
(4\mbox{kg})(4.19\mbox{kJ/kg}\cdot^{\circ}\mbox{C})({t_{f}-90^{\circ}°\mbox{C})=-(16\mbox{kg})(4.19\mbox{kJ/(kg}\cdot^{\circ}\mbox{C})({t_{f}-10^{\circ}\mbox{C})

Solving for t_{f} gives 26°C.

 

[Borek]

I will try it slightly other way - let's assume 10 deg C is a 'basic' state. Your heating adds 4*1000*80 calories. Now you use that heat in the 20000 ml of water. So your delta T is 4*1000*80/20000 = 16 which gives final temperature of 26 and is exactly the same result Sirus posted.


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[chem_tr]

Oookay, so I have made the error of not considering the total volume as Bystander suggested; my logic doesn't seem so wrong. Thank you all!