You mix 50.0 mL of a weak monoprotic acid with 50.0 mL of NaOH solution in a coffee cup calorimeter. Both solutions (and the calorimeter) were initially at 20.5 °C. The final temperature of the neutralization reaction was determined to be 35.7 °C. The calorimeter constant was known to be 21.0 J/°C. (specific heat of H2O = 4.184 J/ g·°C). What is the total amount of heat (in kJ) evolved in this reaction?

6.68kJ

Explanation:

In a neutralization reaction (Of an acid HX and a base as NaOH), some heat is released:

NaOH + HX → H₂O + NaX + ΔH

Where ΔH is heat released.

Now, to find change in heat in a coffee calorimeter you must use the equation:

Q = m×C×ΔT + S×ΔT

Where Q is heat,

m is mass of solution: 100g (50mL of both solutions assuming density of 1g/mL)

C is specific heat of the solution: 4.184J/g°C

ΔT is change in temperature (35.7°C - 20.5°C = 15.2°C)

And S is calorimeter constant: 21.0J/°C

Replacing:

Q = m×C×ΔT + S×ΔT

Q = 100g×4.184J/g°C×15.2°C + 21.0J/°C×15.2°C

Q = 6679J is the heat evolved in the reaction:

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