A Beaker Contains 115 G Of Ethanol At 18.2 C
A baker contains 115 g of ethanol at 18.2 C. What if 1125 J ethanol absorbs heat in the atmosphere without releasing heat? 3
What is the final temperature of ethanol? The specific temperature of ethanol is 2.46 J / g 0 C.
Possible answers are:
A.4.08 C
B.14.1 C
C.18.4 C
D.22.2 C
E.36.4 C
Can someone explain this to me step by step?
Your answer is D
q = mCsà ˆ T (à ˆ T = TfTi)
q = 1125 J m = 115 g Cs = 2.46 J / (g) (Ã) Ti = 18.2 à „ƒ Tf =?
q = mCsà ˆ T
1125J = (115g) [2.46 J / (g) (Ã)] (Tf 18.2 Ã „)
Multiply 115 by 2.46 = 282.9.
Rewrite and cancel.
1125J = [282.9 J / (Ã Â)] (Tf 18.2 Ã „)
Now separate 1125 and 282.9 to cancel 282.9 on the right.
1125J / [282.9 J / (à ƒ)] = Tf 18.2  „
And cancel J.
3977 = Tf 18.2 „
Now add 18.2 بائیں to the left to cancel on the right.
3977 ƒ + 18.2 „ƒ = 22.2
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