Commercial cold packs consist of solid ammonium nitrate and water. NH4NO3 absorbs 330. J of heat per gram dissolved in water. In a coffee-cup calorimeter, 3.00g NH4NO3 is dissolved in 100.0g of water at 24.0 C. What is the final temperature of the solution? Assume that the solution (whose total mass is 103.0g) has a specific heat capacity of 4.18 J/gK.
A) 11.0C
B)15.9C
C)19.1C
D) 21.7C
E)35.9C
The correct answer is D, but when I solve I get a different answer for TF.
so, I found the molar mass of NH4NO3 which is 80.06g/mol
Then I get 3.00g/80.06g per mol , causing grams to cancel, giving a answer of mols which is 0.037471896 mols
Then I use 330J mulitply by what I got from mols which is 0.037471896 mols, giving an answer of 12.36572571J or 12365.72571KJ
we use the formula, q=cmdeltaT
so I plug it in , right lol
12365.72571KJ=(4.18J/gK)(100g)(TF-24.0C)
12365.72571=418(TF-24.0)
then distribute the 418, so it would 418TF- 10032
12365.72571=418TF-10032
So then we have to try to get TF by itself so I move the 10032
10032+12365.72571=418TF
22397.72571=418TF
divide it by 418 to both sides, leaving TF by itself
giving me an answer of 53.5C
which is not answer, so I am wondering what am I doing wrong? Please explian and show correct steps toward the answer.
THANK YOU!!(:
Commercial cold packs consist of solid ammonium nitrate and water. NH4NO3 absorbs 330. J of heat per gram dissolved in water. In a coffee-cup calorimeter, 3.00g NH4NO3 is dissolved in 100.0g of water at 24.0 C. What is the final temperature of the solution? Assume that the solution (whose total mass is 103.0g) has a specific heat capacity of 4.18 J/gK.
Qrxn = -Qsoln
Qrxn = m*HRxn
Qsoln = mass solution*Cp*(Tf-Ti)
then
-m*HRxn = mass solution*Cp*(Tf-Ti)
-(mass)*HRxn = mass solution*Cp*(Tf-Ti)
substitute all data
-(3)*330= 103*4.18*(Tf-24)
Tf = -(3)*330/(103*4.18) + 24
Tf = 21.7°C
NOTE --> you are missing the negativ esign, since reaction losses heat, then the solution "gains") you must add "-" in left side
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