# Finding Specific Heat of 2 substances mixed

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.

## Answers and Replies

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

cdhotfire said:
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.

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

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

cdhotfire said:
A. 2.16
B. 2.095
C. 2.57
D. 4.19
E. 1.0

I got 2.16...

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

and got c= .5147

How did your get 2.16?

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

and got c= .5147

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

I dont understand.

dextercioby
Homework Helper

Daniel.

I did.....

dextercioby
Homework Helper
cdhotfire said:
I dont understand.

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

Daniel.

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

So its

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

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

cdhotfire said:
I obiously dont care what letter it is, I want to know how to do it.

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?

dextercioby
Homework Helper
cdhotfire said:
I obiously dont care what letter it is, I want to know how to do it.

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.

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.

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

dextercioby
Homework Helper
cdhotfire said:
So its

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

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

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.

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

but even still, it ask for the awnser in kilojoules. Wouldnt I multiply it by 1000?

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

dextercioby
Homework Helper
cdhotfire said:
I c how, 1 cal/g*C= 4.184 J/kg*C, correct?

but even still, it ask for the awnser in kilojoules. Wouldnt I multiply it by 1000?

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

Daniel.

Srry for the confusion, I changed my post.

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

Daniel.

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

cdhotfire said:
Srry for the confusion, I changed my post.

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