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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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use enthalpy formula \(q = m \ c\ \Delta t\)
energy lost by hot water = energy absorbed by cool water
How do you set up the entire equation though?

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Other answers:

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

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