Assume you are performing the calibration step of Experiment 8 and you begin with 80 g of water at 20 oC and 80 g of water at 80 oC. After adding the two portions of water into your calorimeter setup and following the procedure outlined in the experiment, you determine the temperature of the mixed portions of water to be 45 oC. What is the heat capacity of the calorimeter?

Assume room temperature is 25 oC.

To determine the heat capacity of the calorimeter, we can use the principle of conservation of energy. The heat lost by the hot water (80 g at 80 oC) is equal to the heat gained by the cold water (80 g at 20 oC) when they are mixed together.

First, let's calculate the heat lost by the hot water using the formula:

Q_lost = m * c * ΔT

Where:
- Q_lost is the heat lost by the hot water
- m is the mass of the hot water (80 g)
- c is the specific heat capacity of water (4.18 J/g°C)
- ΔT is the change in temperature (80 oC - 45 oC = 35 oC)

Q_lost = 80 g * 4.18 J/g°C * 35 oC
Q_lost = 11668 J

Next, let's calculate the heat gained by the cold water using the same formula:

Q_gained = m * c * ΔT

Where:
- Q_gained is the heat gained by the cold water
- m is the mass of the cold water (80 g)
- c is the specific heat capacity of water (4.18 J/g°C)
- ΔT is the change in temperature (45 oC - 20 oC = 25 oC)

Q_gained = 80 g * 4.18 J/g°C * 25 oC
Q_gained = 8360 J

Since energy is conserved, the heat lost by the hot water is equal to the heat gained by the cold water:

Q_lost = Q_gained
11668 J = 8360 J

However, we need to consider that some of the heat is also absorbed by the calorimeter itself. To find the heat capacity of the calorimeter, we can subtract the heat gained by the cold water from the total heat absorbed:

Q_calorimeter = Q_lost - Q_gained
Q_calorimeter = 11668 J - 8360 J
Q_calorimeter = 3308 J

Therefore, the heat capacity of the calorimeter is 3308 J.

To determine the heat capacity of the calorimeter, you need to use the equation:

Q = (M1 + M2) * C * ΔT

Where:
Q is the heat gained or lost by the system
M1 is the mass of the first substance (80 g)
M2 is the mass of the second substance (80 g)
C is the heat capacity of the calorimeter (what we want to find)
ΔT is the change in temperature (difference between the final and initial temperatures, which is 45 oC - 25 oC = 20 oC)

First, calculate the value of Q. Since the system loses heat (water goes from a higher temperature to a lower temperature), Q will be negative:

Q = -Qwater

To calculate Qwater, we can use the equation:

Qwater = M * C * ΔT

Where:
M is the total mass of water (M1 + M2 = 80 g + 80 g = 160 g)
ΔT is the change in temperature (20 oC, as calculated before)

Qwater = 160 g * Cwater * 20 oC

Now, since the heat gained by the water is equal to the heat lost by the calorimeter:

Qwater = -Qcalorimeter
Qwater = -Ccalorimeter * ΔT

Substituting the value of Qwater:

-160 g * Cwater * 20 oC = -Ccalorimeter * ΔT

Now, rearrange the equation to solve for Ccalorimeter:

Ccalorimeter = (160 g * Cwater * 20 oC) / ΔT

Plugging in the values:

Ccalorimeter = (160 g * Cwater * 20 oC) / (45 oC - 25 oC)

Finally, solve the equation to find the heat capacity of the calorimeter.

15 cal/C