Answer to Question #135912 in General Chemistry for Ali

Question

A student mixed 100.00 mL of 1.00 M HCL with 100.0mL of 1.00 M NaOH in a calorimeter flask. The temperature of the HCl before mixing was 24.0 degrees C, and the temperature of the NaOH was 22.0 degrees C. The final temperature of the mixture was 29.0 degrees C. Assuming that the acid-base mixture has a specific heat of 4.184 J/g degrees C, and that the density of this mixture is 1.00 g/mL, compute the heat required to heat this amount of acid-base mixture by the amount recorded by this student.

Answer 1

STEP I: CALCULATE THE INITIAL TEMPERATURE OF THE MIXTURE

Assuming that the mixture were to attain a uniform initial temperature x before the reaction starts;

Then NaOH temp. goes up by (22.0 + x) ^oC

And HCl temp. goes down by (24.0-x) ^oC

22...................x...........................24

Thus 24 to x = 24-x

and x to 22 = x-22

Now since q_lost = q_gain

and q = mC$\Delta$t

Then mC$\Delta$t = mC$\Delta$t

Mass = density x volume

= 1.00g/mL x 200mL

= 200g

Thus;

(200g)(24-x)^oC (4.184J/g^oC) =(200g)(x-22)^oC (4.184J/g^oC)

Solving for x gives

1673.2x = 38492.8 (dividing both sides by 1673.2)

= 23.0^oC (is the initial temperature of the mixture)

STEP II: CALCULATING HEAT CHANGE

Therefore $\Delta$t = (29.0-23.0) = 6.0^oC

q=mC$\Delta$t

q=200g x 4.184J/g^oC x 6^oC

q=5020.8J