anonymous
  • anonymous
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?
Chemistry
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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ganeshie8
  • ganeshie8
use enthalpy formula \(q = m \ c\ \Delta t\)
ganeshie8
  • ganeshie8
energy lost by hot water = energy absorbed by cool water
anonymous
  • anonymous
How do you set up the entire equation though?

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More answers

ganeshie8
  • ganeshie8
.4 * c * (T-25) = -.11 * c * (T-95)
ganeshie8
  • ganeshie8
cancel c, solve T
anonymous
  • anonymous
Got it. Thanks so much!
ganeshie8
  • ganeshie8
yw ! remember this while doing enthalpy related problems : q system = -q surroundings
anonymous
  • anonymous
Actually I have one more question. Why did you use .4 and .11 instead of 400 and 110?
ganeshie8
  • ganeshie8
yeah we can use 400 and 110 also here, it doesnt matter here
ganeshie8
  • ganeshie8
ideally we must convert mL to grams. becoz in enthalpy equation, m represents mass in grams.
ganeshie8
  • ganeshie8
the conversion to grams doesnt matter here since 1 mL = 1 g, for water
ganeshie8
  • ganeshie8
the correct way of putting the equation would be : 400g * c * (T-25) = 110g * c * (T-95)
anonymous
  • anonymous
I ended up getting 19.62 degrees, but it's incorrect :/
ganeshie8
  • ganeshie8
check ur calculation again
ganeshie8
  • ganeshie8
400g * c * (T-25) = -110g * c * (T-95) 400 * (T-25) = -110* (T-95)
ganeshie8
  • ganeshie8
im getting 40.1
anonymous
  • anonymous
That's what I got this time too. Thank you so much! I appreciate all your help.
ganeshie8
  • ganeshie8
u getting 19.62 still ? could u show ur calculation so that i can see whats happening at ur end :)
anonymous
  • anonymous
Nope, I got the correct answer this time. 40.1 degrees Celsius
ganeshie8
  • ganeshie8
sounds great !

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