51.34 g of water at 81.5 oC is added to a calorimeter that contains 52.13 g of water at 40.7 oC. If the final temperature of the system is 59.4 oC, what is the calorimeter constant (Ccalorimeter) ? Use 4.184 J/goC for the heat capacity of water.
[mass H2O x specific heat H2O x (Tfinal-Tinitial)] + [Ccal x (Tfinal-Tinitial)] = 0
To find the calorimeter constant (Ccalorimeter), we can use the principle of energy conservation. The heat gained by the water in the calorimeter is equal to the heat lost by the hot water.
First, let's calculate the heat gained by the colder water:
q1 = m1 * c * ΔT1
Where:
q1 = heat gained by colder water
m1 = mass of colder water
c = specific heat capacity of water
ΔT1 = change in temperature of the colder water
Given:
m1 = 51.34 g
c = 4.184 J/goC (heat capacity of water)
ΔT1 = (final temperature - initial temperature) = 59.4 oC - 40.7 oC
Next, let's calculate the heat lost by the hotter water:
q2 = m2 * c * ΔT2
Where:
q2 = heat lost by hotter water
m2 = mass of hotter water
c = specific heat capacity of water
ΔT2 = change in temperature of the hotter water
Given:
m2 = 52.13 g
c = 4.184 J/goC (heat capacity of water)
ΔT2 = (final temperature - initial temperature) = 59.4 oC - 81.5 oC
According to the principle of energy conservation, q1 = q2. Therefore, we can write:
m1 * c * ΔT1 = m2 * c * ΔT2
Now, we can solve for the calorimeter constant (Ccalorimeter):
Ccalorimeter = q1 / ΔT1
Substituting the values we have:
Ccalorimeter = (m1 * c * ΔT1) / ΔT1
Ccalorimeter = m1 * c
Calculating the value:
Ccalorimeter = 51.34 g * 4.184 J/goC
Ccalorimeter ≈ 214.61 J/oC
Therefore, the calorimeter constant (Ccalorimeter) is approximately 214.61 J/oC.