Question

Please Help...thermochemical equation

How many ML of water at 15 degrees C must be added to 85 ML of water at 65 degrees C to obtain a final temperature of 29 degrees C. ( Assume no heat was lost to surroundings)

Answer 1

@Mertsj can you help?

Answer 2

There is some formula ( I don't remember it) that deals with this. Based on the question, you should have the formula in close proximity.

Answer 3

is it q= m x C x delta T?

Answer 4

yep thats the equation

Answer 5

notice that you need mass and the measurements in the problem are using volume/
so we have to use "density" to do the conversion there

Answer 6

So I think it is that calories lost = calories gained
It takes 1 calorie to raise 1 g of water 1 degree so to lower 85 g of water 36 degrees would require 3060 calories which must be provided by x g of water which will absorb 14x calorie. So I think x = 219 ml

Answer 7

@electrokid 1 ml of water = 1 g of water at 4 degrees celsius

Answer 8

@Mertsj right.
You solved it logically approaching. That aint wrong
But some teachers, want to see that particular equation in there; otherwise, no points. :)

Answer 9

Can you set it up, so i can see it in the order of that equation?

Answer 10

how did you get 36 degrees

Answer 11

Heat lost = Heat gained
\[m_{present}\times c\times (65-29)=m_{added}\times c \times (29-15)\]

Answer 12

\[\cancel{\rho}\times85\times\cancel{c}\times 34=\cancel{rho}\times v\times\cancel{c}\times 14\\
v={85\times34\over14}=?{\rm ml}
\]

Answer 13

So do i have to do anything with the heat lost= heat gained?

Answer 14

thats how I got the equations

Answer 15

where did you get the 34

Answer 16

look at the step right before that.

Answer 17

found it?

Answer 18

65-29=36 is that it?

Answer 19

oh. my bad. yea.. 36 :)

Answer 20

o ok so then it would be 218.6

Answer 21

right. just like what @Mertsj got

Answer 22

ok thank you

Answer 23

yw.