Question

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?

Answer 1

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

Answer 2

energy lost by hot water = energy absorbed by cool water

Answer 3

How do you set up the entire equation though?

Answer 4

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

Answer 5

cancel c, solve T

Answer 6

Got it. Thanks so much!

Answer 7

yw !

remember this while doing enthalpy related problems :

q system = -q surroundings

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

the correct way of putting the equation would be :

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

Answer 13

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

Answer 14

check ur calculation again

Answer 15

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

Answer 16

im getting 40.1

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

sounds great !