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