anonymous
  • anonymous
If you have 430.0 mL of water at 25.00 °C and add 140.0 mL of water at 95.00 °C, what is the final temperature of the mixture? Use 1.00 g/mL as the density of water.
Chemistry
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
I just need help with an equation to use. i've looked over all my notes and can't seem to come up with anything. I don't really want an answer, more of a bit of help.
anonymous
  • anonymous
you will also need heat capacity of water... and then you go with this equation: Q = m * cp * (T2-T1) where Q = heat m = mass cp = heat capacity of water (T2-T1) = temperature now since you have heating of colder water and cooling of hotter water you write: Q(hot) = - Q(cold) and T2 is your final temperature on both sides so you have: m(h) * cp(w) * (T2 - 95) = m(c) * cp(w) * (T2 - 25) where indexes are: h - hot c - cold w - water and all you need to do is to solve that equation for T2 and you will get your answer! hope this helps ;)

Looking for something else?

Not the answer you are looking for? Search for more explanations.