## allikatt 3 years ago 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?

1. ganeshie8

use enthalpy formula \(q = m \ c\ \Delta t\)

2. ganeshie8

energy lost by hot water = energy absorbed by cool water

3. allikatt

How do you set up the entire equation though?

4. ganeshie8

.4 * c * (T-25) = -.11 * c * (T-95)

5. ganeshie8

cancel c, solve T

6. allikatt

Got it. Thanks so much!

7. ganeshie8

yw ! remember this while doing enthalpy related problems : q system = -q surroundings

8. allikatt

Actually I have one more question. Why did you use .4 and .11 instead of 400 and 110?

9. ganeshie8

yeah we can use 400 and 110 also here, it doesnt matter here

10. ganeshie8

ideally we must convert mL to grams. becoz in enthalpy equation, m represents mass in grams.

11. ganeshie8

the conversion to grams doesnt matter here since 1 mL = 1 g, for water

12. ganeshie8

the correct way of putting the equation would be : 400g * c * (T-25) = 110g * c * (T-95)

13. allikatt

I ended up getting 19.62 degrees, but it's incorrect :/

14. ganeshie8

check ur calculation again

15. ganeshie8

400g * c * (T-25) = -110g * c * (T-95) 400 * (T-25) = -110* (T-95)

16. ganeshie8

im getting 40.1

17. allikatt

That's what I got this time too. Thank you so much! I appreciate all your help.

18. ganeshie8

u getting 19.62 still ? could u show ur calculation so that i can see whats happening at ur end :)

19. allikatt

Nope, I got the correct answer this time. 40.1 degrees Celsius

20. ganeshie8

sounds great !