When 50.0mL of .10M HCl and 50.0mL of .10NaOH, both at 22C are added to a calorimeter,the temperature of the mixture reaches 28.9C. Calculate the heat produced by this reaction. Density of water 1.00g/mL specific heat water=4.184J/g C
well, 100ml increased temp by 6.9C.
heat=mc deltatemp=100g*4.184* 6.8 C=..... Joules
To calculate the heat produced by this reaction, you can use the equation:
q = m * c * ΔT
Where:
q is the heat produced (in joules)
m is the mass of the solution (in grams)
c is the specific heat of water (in J/g°C)
ΔT is the change in temperature (in °C)
1. First, let's calculate the mass of the solution. Since the density of water is 1.00 g/mL, the volume of the solution is 50.0 mL + 50.0 mL = 100.0 mL. Therefore, the mass of the solution is 100.0 g (because 100.0 mL = 100.0 g).
2. Next, let's calculate the change in temperature (ΔT). ΔT = 28.9°C - 22°C = 6.9°C.
3. Now, substitute the values into the equation:
q = 100.0 g * 4.184 J/g°C * 6.9°C
4. Calculate the heat produced:
q = 2866.36 J
Therefore, the heat produced by this reaction is 2866.36 J.