Question

In the following experiment, a coffee-cup calorimeter containing 100 g of H2O is used. The initial temperature of the calorimeter is 23.0 oC. If 3.3 g of CaCl2is added to the calorimeter, what will be the final temperature of the solution in the calorimeter? Assume that the heat capacity of the solution is 4.18 J/goC, and that the heat capacity of the calorimeter is negligible. The heat of dissolution ΔHsoln of CaCl2 is −82.8 kJ/mol.

Answer 1

Answer:

The final temperature of the solution in the calorimeter is 28,7°C

Explanation:

A -ΔH means that when you dissolve CaCl₂ you will produce heat. This heat you produce is:

3,3g CaCl₂×$\frac{1 molCaCl_{2}}{110,98g}$×82800J/mol = 2462,1 J

The specific heat capacity of solution is:

q = c×m×ΔT

Where q is energy: (2462,1J)

c is heat capacity of the solution (4,18J/g°C)

m is mass of solution (103,3g)

And ΔT is (T-23°C)

Solving for T:

$T = 23C+\frac{2462,1J}{4,18J/gC*103,3g}$

T = 28,7°C

I hope it helps!