Calculate ΔHrxn for Liquid -> Solid Water at -5°C

[Neophyte]

Homework Statement

What is the heat change in J associated with 80.7 g of liquid water at 5.00 ° C changing to solid water at -5.00 °C?

Homework Equations

c(H2 O) (liq) = 4.184 J/(g.K)

c(H2O) (s) = 2.09 J/ (g.K)

DHfus(H2 O) = 6.02 kJ/mol

The Attempt at a Solution

Δt = 5°

While liquid
(80.7g)(4.184J/g.k)(5°) = 1688.2 J
(-6.02kJ/mol)(80.7/18 = 4.48 mol) = -27.0 kJ
(80.7g)(2.09J/g.k)(5°)= 843.3 J

2531.5 - 27000
-2.44E4
(correct answer is -2.95E-4)

I mean I was pretty sure that is how you went about doing this type of problem :/.
The Δt = 5° came from (5 - 0) (and 0-(-5))

 

[chemisttree]

I believe you are messing up in your sign convention. You are removing heat throughout this process and thus all of your signs must be negative. Delta T is -5, not 5.

 

[Blueshift5]

so there would be no negative degrees. The molar mass of H2O is 18 g/mol, so 80.7 g is equivalent to 80.7/18 = 4.48 mol.

Dear student,

Thank you for your response. Your calculations are correct, but your final answer is incorrect. The correct answer is -2.95 x 10^-4 J, not -2.44 x 10^-4 J. This is because you made a minor error in your last step of subtraction. The correct calculation is 2531.5 J - 27000 J = -2.95 x 10^-4 J.

It is important to pay attention to significant figures when doing calculations. In this case, the given values have three significant figures, so the final answer should also have three significant figures. Therefore, the final answer should be -2.95 x 10^-4 J.

I hope this helps clarify any confusion. Keep up the good work in your studies!