? Vol. of water @ 18.2*C is added to 23.2ml of water @35.0*C. If final temp is 23.5what was unknown Vol.?

Question

assume no heat is lost to surroundings ; d of water is 1.00g/ml
I would appreciaste any help learning how to work these problems, even with a different prob set up thank you, I have a serious problem with setting these up

Answer 1

This is actually just bookkeeping.

In all these heat transfer problems, the heat gained by the colder object is equal to the heat lost by the warmer object (ignoring losses to the surroundings).

And, by the definition of specific heat,

Heat gained (or lost) = mass x specific heat x increase (or decrease) in temperature

So in this case,

? ml water x 1 g/ml x (23.5 - 18.2) x specific heat of water

which is the heat gained by the colder object, must be equal to

23.2 ml water x 1g/ml x (35.0 - 23.5) x specific heat of water

So you have a simple equation, there is only one thing you don't know, and you can solve it.

There are lots of variants on this, but they all work the same way; there is only one thing that you are not told and you have to put heat loss equal to heat gained, plug in the information that you do know, and solve for what you don't.

Good luck!