Step-by-step explanation:
a) Sodium Hydroxide:
Volume of water = 100.0 mL
Initial temperature of water = 21.3 °C
Mass of NaOH = 9.0 g
Final temperature of water = 41.3 °C
So we had a 100.0 mL sample of water at 21.3 °C and after we added 9.0 g of NaOH the temperature increased to 41.3 °C. Let's find the heat that the water absorbed using the formula:
Q = m * Cp * ΔT
Where m is the mass, Cp is the specific heat of water (Cp = 4.186 J/g°C) and ΔT is the temperature change.
First we had 100.0 mL of water or 100.0 g (if we consider that the density of water is 1.0 g/mL) and we added 9.0 g of NaOH. Let's find the total mass.
m = total mass = mass of water + mass of NaOH
m = 100.0 mL * 1.0 g/mL + 9.0 g
m = 109.0 g
We can also find the change in temperature.
ΔT = Tfinal - Tinitial = 41.3 °C - 21.3 °C
ΔT = 20.0 °C
The dissolution of the NaOH is an exothermic process. This reaction will release heat that will be absorbed by its surroundings and the temperature of the water will increase.
Finally we can calculate the heat that was absorbed.
Qw = m * Cp * ΔT
Qw = 109 g * 4.186 J/(g°C) * 20.0 °C
Qw = 9125 J
So the water absorbed 9125 J and the reaction released -9125 J.
b) Sodium thiosulfate pentahydrate:
Volume of water = 100.0 mL
Initial temperature of water = 21.3 °C
Mass of sodium thiosulfate pentahydrate = 20.0 g
Final temperature of water = 14.2 °C
The dissolution of the sodium thiosulfate pentahydrate is an endothermic process in this case. The dissolution will absorb energy from its surroundings and the temperature of the water will decrease.
m = total mass = mass of water + mass of thiosulfate
m = 100.0 mL * 1.0 g/mL + 20.0 g
m = 120.0 g
ΔT = Tfinal - Tinitial = 14.2 °C - 21.3 °C
ΔT = -7.1 °C
Qw = m * Cp * ΔT
Qw = 120 g * 4.186 J/(g°C) * (-7.1 °C)
Qw = -3566 J
So the water released -3566 J and the dissolution process absorbed 3566 J.
Answer:
a) The water absorbed 9125 J and the dissolution process released -9125 J.
b) The water released -3566 J and the dissolution process absorbed 3566 J.