# Homework Help: Finding Specific Heat of 2 substances mixed

1. Jan 8, 2005

### cdhotfire

In an experiment, .02 kg of substance is heated in a bath of boiling water until its temperature is apporoximately 100*C. The substance is the placed in an insulated container that is holding .10 kg of water at 25*C. The final equilibrium temperature of the mixture is 32*C. The specific heat of substance, in kilojoules kilogram . degree Celsius, is.

I tried making the change of heat of the substance equal to change of heat of wate, but that didnt give me the right awnser, I know this because it is a multiple choice awnser. I would really appreciate help on this. Thxs before hand.

2. Jan 8, 2005

### cdhotfire

Oops, that title is so wrong, suppose to be, Finding Specific Heat of Substance. What was I thinking.

3. Jan 8, 2005

### apchemstudent

That should work... what are the answers available?

4. Jan 8, 2005

A. 2.16
B. 2.095
C. 2.57
D. 4.19
E. 1.0

5. Jan 8, 2005

### apchemstudent

I got 2.16...

6. Jan 8, 2005

### cdhotfire

I did:
(.02 kg)(68*C)c=(.1 kg)(1)(7*C)

and got c= .5147

7. Jan 8, 2005

### cdhotfire

How did your get 2.16?

8. Jan 8, 2005

### apchemstudent

you got the answer in calories/(K*kg) multiply it by the specific heat of water in KJ

9. Jan 8, 2005

### cdhotfire

I dont understand.

10. Jan 8, 2005

### dextercioby

Daniel.

11. Jan 8, 2005

### cdhotfire

I did.....

12. Jan 8, 2005

### dextercioby

That "1" in the formula for heat absorbed by water.It's 1 calory.It should be in $KJ Kg^{-1}K^{-1}$

Daniel.

13. Jan 8, 2005

### apchemstudent

For the specific heat of water, use 4.19kJ/(K*kg) instead of 1 calorie/(K*kg)

14. Jan 8, 2005

### cdhotfire

So its

(.02 kg)(68*C)c=(.1)(7*C)(1 ?)

what is the specific heat measured in, isn't cal/g*C?

15. Jan 8, 2005

### apchemstudent

If you actually read between the line... you would notice that we've actually told you how to do it...

Dextercioby, should we consider helping someone this rude?

16. Jan 8, 2005

### dextercioby

Okay,u did it fine,but put wrong units.If u put instead of calories $KJ Kg^{-1}K^{-1}$,your answer will coincide with the one at point "a".

Daniel.

17. Jan 8, 2005

### cdhotfire

apchem, I posted that before, because, dex said i got A, and told me to post my work. I said I did post it, and I didnt really care what the letter awnser was.

Anyways thxs for the help, didnt mean to be rude.

18. Jan 8, 2005

### apchemstudent

Use the 4.19 in the place of 1.... since that is the correct unit not 1 calorie/(K*kg)

19. Jan 8, 2005

### dextercioby

Yes,it's just that u need to express it in KJ/Kg*C (or Kelvin,they are the same thing,just scaling is different).

Daniel.

20. Jan 8, 2005

### cdhotfire

I c how, 1 cal/g*C= 4.184 J/kg*C, correct?

Oops, 4.184 is already in KJ, I just noticed, that 1 cal/g*C=4184 J/kg*C. Thats whats started this whole question.

Well thxs for all the help, and again I didn't mean to be rude.

Last edited: Jan 8, 2005
21. Jan 8, 2005

### apchemstudent

nahh... 1 calorie = 4.184 J.... so multiply the answer by 4.184 and see what you get...

22. Jan 8, 2005

### dextercioby

Yes u would have to multiply it.Else your answer would be in J/Kg*C instead of KJ/Kg*C

Daniel.

23. Jan 8, 2005

### cdhotfire

Srry for the confusion, I changed my post.

24. Jan 8, 2005

### apchemstudent

first of all Its 4.184J/g*K(orC)

and if you convert it to 4.184kJ/kg*K(orC)

you don't need to multiply by anything. so thus... only need to muliply by 4.184

25. Jan 8, 2005

### apchemstudent

I see your change... well then you will have to divide by 1000 and multiply by 4184 to your original answer of 0.516