Question

In the following experiment, a coffee-cup calorimeter containing 100 mL of H2O is used. The initial temperature of the calorimeter is 23 degrees Celsius. If 4.00 g of CaCl2 is added to the calorimeter, what will be the final temperature of the solution in the calorimeter? The heat of solution of CaCl2 is -82.8 kJ/mol.

Answer 1

Here are my numbers:
100mL water, Temp 1: 23C, Temp2: ?, 7.20 g CaCl2, DeltaH=-82.8kJ/mol.

First:
Temperature is going to increase, so turn your -82.8kJ/mol to +82.8.
Also, 100 mL water = 100 g water b/c water's density is 1g/mL

Second:
82.8kJ/mol=82,800J/mol

Third:
Molar mass of CaCl2=110.9 g/mol

Fourth:
(82,800 J/mol)(1mol/110.98g)=746.08 J/g

Fifth:
746.08J/g * 7.20 g = 5371.78 J

Sixth:
(100g + 7.2 g total mass)(4.18 specific heat)(T2-23C)=5371.78J

T2-23=12
T2=35C

HOpe this helps!

Answer 2

Wow finally someone who actually was able to help me get the right answer!!! ^ (I'm not sure where you got 7.2g for CaCl2, but apart from that you did everything right!)

Thanks for the help!

Answer 3

^^ yes this is how you get to the right answer, if you were given a dif temp just plug it in and solve.