In an experiment to determine heat loss to a calorimeter, 100.0 mL of cool water was measured into the calorimeter and allowed to reach room temperature before 100.0 mL of heated water was added. The results of this experiment produced the data below. What is the absolute value of the calorimeter constant in J / ºC?
ΔTheated water = -6.2 ºC
ΔTcool water= 11.2 ºC
Specific heat of water = 4.184 J deg-1 g-1
Density of water = 1 g / mL
Mass of calorimeter = 40.91 g
I am confused by the wording in the problem. For example, let's say room T is 10 C so we put 100 mL of cool water in the calorimeter and the final T is 3.8 to make dT = -6.2. Then we allow the system to warm back up to 10 C? Is that right?
To find the absolute value of the calorimeter constant, we first need to calculate the heat gained by the cool water and the heat lost by the heated water.
The heat gained or lost by a substance can be calculated using the formula:
Q = mcΔT
Where:
Q = heat gained or lost (in Joules)
m = mass of the substance (in grams)
c = specific heat capacity of the substance (in Joules per gram per degree Celsius)
ΔT = change in temperature (in degrees Celsius)
For the cool water:
mass = volume x density = 100.0 mL x 1 g/mL = 100.0 g
ΔT = 11.2 ºC
Qcool water = mcΔT
Qcool water = 100.0 g x 4.184 J/g°C x 11.2 ºC = 4647.04 J
For the heated water:
mass = volume x density = 100.0 mL x 1 g/mL = 100.0 g
ΔT = -6.2 ºC
Qheated water = mcΔT
Qheated water = 100.0 g x 4.184 J/g°C x -6.2 ºC = -2595.232 J
Next, we need to calculate the heat lost by the calorimeter. Since the cool water and heated water reached equilibrium at room temperature, the heat lost by the heated water must have been gained by the calorimeter and the cool water.
Qcalorimeter = Qcool water + Qheated water
Qcalorimeter = 4647.04 J - 2595.232 J = 2051.808 J
Finally, we can calculate the calorimeter constant by dividing Qcalorimeter by the change in temperature of the calorimeter.
ΔTcalorimeter = ΔTcool water = 11.2 ºC
Calorimeter constant = Qcalorimeter / ΔTcalorimeter
Calorimeter constant = 2051.808 J / 11.2 ºC
The absolute value of the calorimeter constant in J/ºC is the magnitude of this result.
To determine the calorimeter constant, we need to use the equation:
q = mcΔT
Where:
q is the heat gained or lost by the water and the calorimeter (in this case, gained by the water and lost by the calorimeter)
m is the mass of the water (in this case, the sum of the masses of the heated and cool waters)
c is the specific heat of water
ΔT is the change in temperature of the water
We can calculate the value for q using the equation for each water sample, and then add them together to get the total heat gained by the water. Since the calorimeter is losing heat, we'll take the negative sign into account.
For the heated water:
m = density × volume = 1 g/mL × 100 mL = 100 g
ΔT = -6.2 ºC
c = 4.184 J deg-1 g-1
q1 = mcΔT = 100 g × 4.184 J deg-1 g-1 × (-6.2 ºC) = -2599.648 J
For the cool water:
m = density × volume = 1 g/mL × 100 mL = 100 g
ΔT = 11.2 ºC
c = 4.184 J deg-1 g-1
q2 = mcΔT = 100 g × 4.184 J deg-1 g-1 × 11.2 ºC = 4666.048 J
Now, to find the calorimeter constant, we can rearrange the equation to solve for it:
q1 + q2 = CΔT
Where:
C is the calorimeter constant we want to find
ΔT is the difference in temperature between the water and the surroundings (room temperature)
Rearranging the equation:
C = (q1 + q2) / ΔT
Since each q value is in joules and ΔT is in degrees Celsius, the resulting calorimeter constant will be in joules per degree Celsius.
C = (-2599.648 J + 4666.048 J) / ΔT
Let's assume the difference in temperature between the water and the surroundings (room temperature) is 20 ºC.
C = (2066.4 J) / (20 ºC)
C = 103.32 J / ºC
So, the absolute value of the calorimeter constant is 103.32 J / ºC.