Question

Please help Ill give a medal.
What is the initial temperature of 13 grams of water if it ends up at 25 degrees C and the water releases 1250 calories?

Answer 1

First question: if the water releases 1250 calories, is the water getting warmer, or cooler?

Answer 2

cooler

Answer 3

Good. And do you think it is cooling down enough that we have to worry about a change between phases or states?

Answer 4

umm yes I think so

Answer 5

We'll find out :-)

Answer 6

Haha okay

Answer 7

Do you have any formulas and constants at hand for this problem?

Answer 8

I think were supposed to use Q= C*m*deltaT

Answer 9

\[Q = C m \Delta T\]\[C = \text{specific heat of water}\]\[\Delta T = \text{change in temperature}\]\[m = \text{mass of water} = 13 \text{ g}\]\[Q = -1350 \text{ cal}\]

Do you know the specific heat of water?

Answer 10

Yeah its 1 correct?

Answer 11

wait why does Q= -1350 cal? did you mean -1250 cal?

Answer 12

Yes.

Sorry, yes, hit the wrong key, good catch!

\[1250 = (1)(13)(T_i - 25)\]if \(T_i\) is the initial temperature of the water

Can you solve that for \(T_i\)?

Answer 13

Okay I got 96.15=(Ti-25)
Would I divide by -25 now?

Answer 14

No, \[96.15 = T_i - 25\]Add 25 to each side:
\[96.15+25 = T_i \cancel{- 25} \cancel{+ 25}\]\[121.15 = T_i\]

Answer 15

Dividing by -25 would have given us
\[\frac{96.15}{-25} = \frac{T_i}{-25} -\frac{25}{-25}\]\[-3.846 = -\frac{T_i}{25} +1\]Which isn't really any closer to the solution :-)

Answer 16

ohhhh I didnt think of doing that ...Thank you you've been a big help (:

Answer 17

Want to help me with one more? Haha

Answer 18

I think this one's easier I just cant figure out what to do

Answer 19

Now, I'm a little concerned that that means the initial temperature was apparently above the boiling point of water at standard temperature and pressure.

Answer 20

So we did it wrong?

Answer 21

Well, if it was in the form of liquid water, somehow prevented from boiling into steam, then we did it right. But I could imagine a problem where we were supposed to realize that it started out as steam, lost heat getting down to the boiling point, lost (much) more heat changing to liquid water, and then cooled down to 25 C or whatever it was.

Answer 22

have you done any problems like that in class yet?

Answer 23

No he just gives us practice sheets and some of them we have to figure out on our own...I think we did it correct though.

Answer 24

Okay. I took a look at it assuming that we did have to do a state change, and either I did it wrong (might have used the wrong constant), or the water was partway through the change already, which doesn't seem like how I would expect the problem to be written, so I'll just go with what we've got if you're comfortable with that. Sounds like you are.

Answer 25

Just be prepared to encounter such problems in the near future, I guess :-)

You had another problem?

Answer 26

Think you could help me with this one? :P
If the specific heat capacity of water is 1 cal/(gdegrees C), how much heat is needed to change the temperature of 150 grams of water by 15 degrees C?

Answer 27

And yeah I feel good with the answer because I don't think he would give us anything too crazy

Answer 28

\[Q = mC\Delta T\]\[Q = (150 \text{ g})(1 \frac{\text{ cal}}{\text{g}^\circ\text{C}})(15^\circ\text{C}) = (150 \cancel{\text{ g}})(1 \frac{\text{ cal}}{\cancel{\text{g}}\cancel{^\circ\text{C}}})(15\cancel{^\circ\text{C}}) = 150*1*15 \text{ cal}\]

Answer 29

So just multiplying those together will give me the answer?

Answer 30

I believe so, yes.

Answer 31

That's what I did from the start but I thought I wasn't right... I guess I just needed some reassurance. Thank you (:

Answer 32

If it turns out to be wrong, contact me and I'll cheerfully refund you everything you've paid me :-)

Answer 33

Hahaha I would pay you if I could because I would've gotten these wrong if it weren't for your help (:

Answer 34

Hey on the first one isn't the specific heat of water 4.18 j/g*k

Answer 35

Oh wait sorry I just realized it was in cal not j