50.00 mL of 1.0 M HCl and 50.00 mL of 1.0 M NaOH are mixed. The temperature is observed to increase by 10.0 degrees C. How much energy (in J) was absorbed by the calorimeter? Assume a density of the reaction mixture of 1 g/mL. Specific heat of the solution is 5.00 J/g/degree C. The calorimeter constant is 10.0 J/degree C.
The energy absorbed by the calorimeter itself is constant x delta T = 10 x 10.
The energy absorbed by the calorimeter AND the contents is
[mass soln x specific heat soln x delta T] + [Ccal x delta T}= ??
To calculate the energy absorbed by the calorimeter, we need to determine the heat transferred during the reaction. The heat transferred can be calculated using the formula:
q = m * c * ΔT
Where:
q - heat transferred
m - mass of the solution
c - specific heat of the solution
ΔT - change in temperature
First, we need to determine the mass of the solution. Since the density of the reaction mixture is given as 1 g/mL and the total volume of the solution is 100.00 mL, the mass can be calculated as:
mass = volume * density
mass = 100.00 mL * 1 g/mL
mass = 100.00 g
Now, we can calculate the heat transferred using the formula. The specific heat of the solution is given as 5.00 J/g/degree C, and the change in temperature is given as 10.0 degrees C. Therefore:
q = 100.00 g * 5.00 J/g/degree C * 10.0 degrees C
q = 5000 J
The heat transferred, q, is equal to the energy absorbed by the calorimeter. However, we need to account for the calorimeter constant, which is given as 10.0 J/degree C. Since the temperature change is 10.0 degrees C, the energy absorbed by the calorimeter can be calculated as:
Energy absorbed by the calorimeter = q - (calorimeter constant * ΔT)
Energy absorbed by the calorimeter = 5000 J - (10.0 J/degree C * 10.0 degrees C)
Energy absorbed by the calorimeter = 5000 J - 100 J
Energy absorbed by the calorimeter = 4900 J
Therefore, 4900 J of energy was absorbed by the calorimeter.