You mix 50 ml of a weak monoprotic acid with 50 ml of NaOH solution in a coffee

Question

You mix 50 ml of a weak monoprotic acid with 50 ml of NaOH solution in a coffee cup calorimeter. Both solutions and the calorimeter were initially at 22 C. The final temperature of the neutralization reaction was determined to be 33.5 C. The colorimeter constant was kown to be 20.6 J/C. Specific heat of water is 4.18

a. What is the total amount of heat evolved in this reaction

b. if 135 mmols of the monoprotic acid were neutralized in this reaction, what is the molar heat of neutralization for this reaction.

Explanation / Answer

The rule of heat states:
energy=mass*specific*change in temp.
E=mc(t2-ti)

m=? c=4.18 t2=33.5 t1=22
E=?

so the mass needs to be calculated first......u mixed 50mL with another 50mL of solutions to get 100mL = 100cm^3 = 100g as density = 1 of these solutions like the water
**actually the density isn't exactly 1 but close to it**

so m=100 c=4.18 t2=33.5 t1=22
so total energy=100*4.18*(33.5-22)
=4807 j =4.8kj

i guess but not sure the calorimeter constant tries to tell u neglect it..heat doesn't affect the calorimeter.

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