Suppose 150.0 mL of 0.50 M HCl and 150.0 mL of 0.50 M NaOH, both initially at 24.0°C, are mixed in a thermos flask. When the reaction is complete, the temperature is 27.4°C. Assuming that the solutions have the same heat capacity as pure water, compute the heat released (in kJ).
q = mass water x specific heat water x (Tfinal-Tinitial).
mass water is 300 mL = 300 g
To solve this problem, we can use the equation for heat released in a reaction:
q = m * C * ∆T
where:
q is the heat released (in joules)
m is the mass of the solution (in grams)
C is the specific heat capacity of water (4.18 J/g°C)
∆T is the change in temperature (in °C)
First, let's calculate the mass of the solution. Since we have equal volumes of both HCl and NaOH, the total volume of the solution is 150.0 mL + 150.0 mL = 300.0 mL.
Next, we need to convert the volume to grams using the density of water. The density of water is 1.00 g/mL, so the mass of the solution is:
mass = volume * density
mass = 300.0 mL * 1.00 g/mL
mass = 300.0 g
Now, let's calculate the change in temperature (∆T):
∆T = final temperature - initial temperature
∆T = 27.4°C - 24.0°C
∆T = 3.4°C
Finally, let's plug the values into the equation to calculate the heat released (q):
q = 300.0 g * 4.18 J/g°C * 3.4°C
q = 4236 J
Since the question asks for the heat released in kJ, we divide by 1000 to convert joules to kilojoules:
q = 4236 J / 1000
q = 4.236 kJ
Therefore, the heat released during the reaction is 4.236 kJ.