allikatt
If you have 400.0 mL of water at 25.00 °C and add 110.0 mL of water at 95.00 °C, what is the final temperature of the mixture?



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ganeshie8
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use enthalpy formula \(q = m \ c\ \Delta t\)

ganeshie8
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energy lost by hot water = energy absorbed by cool water

allikatt
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How do you set up the entire equation though?

ganeshie8
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.4 * c * (T25) = .11 * c * (T95)

ganeshie8
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cancel c, solve T

allikatt
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Got it. Thanks so much!

ganeshie8
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yw !
remember this while doing enthalpy related problems :
q system = q surroundings

allikatt
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Actually I have one more question. Why did you use .4 and .11 instead of 400 and 110?

ganeshie8
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yeah we can use 400 and 110 also here, it doesnt matter here

ganeshie8
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ideally we must convert mL to grams. becoz in enthalpy equation, m represents mass in grams.

ganeshie8
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the conversion to grams doesnt matter here since 1 mL = 1 g, for water

ganeshie8
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the correct way of putting the equation would be :
400g * c * (T25) = 110g * c * (T95)

allikatt
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I ended up getting 19.62 degrees, but it's incorrect :/

ganeshie8
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check ur calculation again

ganeshie8
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400g * c * (T25) = 110g * c * (T95)
400 * (T25) = 110* (T95)

ganeshie8
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im getting 40.1

allikatt
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That's what I got this time too. Thank you so much! I appreciate all your help.

ganeshie8
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u getting 19.62 still ? could u show ur calculation so that i can see whats happening at ur end :)

allikatt
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Nope, I got the correct answer this time. 40.1 degrees Celsius

ganeshie8
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sounds great !