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allikatt
Group Title
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?
 2 years ago
 2 years ago
allikatt Group Title
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?
 2 years ago
 2 years ago

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ganeshie8 Group TitleBest ResponseYou've already chosen the best response.2
use enthalpy formula \(q = m \ c\ \Delta t\)
 2 years ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.2
energy lost by hot water = energy absorbed by cool water
 2 years ago

allikatt Group TitleBest ResponseYou've already chosen the best response.0
How do you set up the entire equation though?
 2 years ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.2
.4 * c * (T25) = .11 * c * (T95)
 2 years ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.2
cancel c, solve T
 2 years ago

allikatt Group TitleBest ResponseYou've already chosen the best response.0
Got it. Thanks so much!
 2 years ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.2
yw ! remember this while doing enthalpy related problems : q system = q surroundings
 2 years ago

allikatt Group TitleBest ResponseYou've already chosen the best response.0
Actually I have one more question. Why did you use .4 and .11 instead of 400 and 110?
 2 years ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.2
yeah we can use 400 and 110 also here, it doesnt matter here
 2 years ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.2
ideally we must convert mL to grams. becoz in enthalpy equation, m represents mass in grams.
 2 years ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.2
the conversion to grams doesnt matter here since 1 mL = 1 g, for water
 2 years ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.2
the correct way of putting the equation would be : 400g * c * (T25) = 110g * c * (T95)
 2 years ago

allikatt Group TitleBest ResponseYou've already chosen the best response.0
I ended up getting 19.62 degrees, but it's incorrect :/
 2 years ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.2
check ur calculation again
 2 years ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.2
400g * c * (T25) = 110g * c * (T95) 400 * (T25) = 110* (T95)
 2 years ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.2
im getting 40.1
 2 years ago

allikatt Group TitleBest ResponseYou've already chosen the best response.0
That's what I got this time too. Thank you so much! I appreciate all your help.
 2 years ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.2
u getting 19.62 still ? could u show ur calculation so that i can see whats happening at ur end :)
 2 years ago

allikatt Group TitleBest ResponseYou've already chosen the best response.0
Nope, I got the correct answer this time. 40.1 degrees Celsius
 2 years ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.2
sounds great !
 2 years ago
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