The temperature increase when 100 mL of acid and 100 mL of base had been used in the same calorimeter is the same starting at the same temperature of 22.0 °C is 3.45 degrees C.
The quantity of heat involved in a given reaction is given by the formula below:
Qsolution = (c × m × ΔT)
The quantity of heat involved in the reaction of 50 mL of acid and 50 mL of base described in Example 5.5 is given below:
Q = (4.184 J/g °C)(1.0 × 10²g)(28.9 °C − 22.0 °C)
Q = 2.89 × 10³ J
The quantity of heat involved when 100 mL of acid and 100 mL of base had been used in the same calorimeter is the same.
The mass of water has been doubled from 100 g to 200 g.
In order for the heat involved to stay the same, the temperature increase has to be halved.
Temperature increase = 6.9/2 = 3.45 degrees C
In conclusion, the quantity of heat involved in the reaction is the heat of neutralization.
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