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

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In summary, the forum members say that if we heat 4 liters of water from 10°C to 90°C, and add this to 16 liters of water at 10°C, the final temperature will be 30°C.
  • #1
chem_tr
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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 [tex]Q=m \times c \times \Delta t=4000 \times 1 \times (90-10)=320000~cal[/tex]. This amount of energy is dissipated in the whole pool of water, so [tex]320000=16000 \times 1 \times (t_{final}-10)=>t_{final}[/tex] 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.
 
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  • #2
You have a total of 20 liters, not 16. That help?
 
  • #3
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?

[tex]mc(x-90)=m_2c(x-10)[/tex]

-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
 
  • #4
Let me try:

[tex](\mbox{heat lost})=-(\mbox{heat gained})[/tex]
[tex](mc\Delta t)_{1}=-(mc\Delta t)_{2}[/tex]
[tex](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})[/tex]

Solving for [itex]t_{f}[/itex] gives 26°C.
 
  • #5
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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  • #6
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!
 

What is calorimetry?

Calorimetry is the scientific measurement of the amount of heat energy released or absorbed during a chemical reaction or physical change.

What is the purpose of using calorimetry in scientific experiments?

Calorimetry is used to determine the specific heat capacity, enthalpy, and other thermodynamic properties of substances. It is also used to measure the energy content and nutritional value of food.

How does a calorimeter work?

A calorimeter works by measuring the change in temperature of a known mass of substance when it undergoes a chemical reaction or physical change. This change in temperature is used to calculate the amount of heat released or absorbed.

What are the different types of calorimeters?

There are two main types of calorimeters: bomb calorimeters and differential scanning calorimeters. Bomb calorimeters are used for measuring the heat of combustion, while differential scanning calorimeters are used for measuring thermal properties such as melting point and glass transition temperature.

What are some real-world applications of calorimetry?

Calorimetry is used in a variety of industries, including medicine, food science, and environmental science. It is used to measure the nutritional value of food, determine the effectiveness of medicines, and study the impact of climate change on ecosystems.

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