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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?
 one year ago
 one year 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?
 one year ago
 one year ago

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

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

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

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

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

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

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

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

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

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

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.2
im getting 40.1
 one year 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.
 one year 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 :)
 one year ago

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

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