Assume you are performing the calibration step of Experiment 8 and you begin with 70 g of water at 20 oC and 70 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.

Indicate your subject in the "School Subject" box, so those with expertise in the area will respond to the question.

To determine the heat capacity of the calorimeter, we can use the formula:

q(calorimeter) = -q(water)

where q(calorimeter) is the heat gained or lost by the calorimeter and q(water) is the heat gained or lost by the water.

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

q(water) = m(water) * c(water) * ΔT

where m(water) is the mass of water, c(water) is the specific heat capacity of water, and ΔT is the change in temperature of the water.

For the hot water:

m(hot) = 70 g
c(water) = 4.18 J/g°C (specific heat capacity of water)
ΔT(hot) = 45°C - 80°C = -35°C (negative because the temperature is decreasing)

q(hot) = 70 g * 4.18 J/g°C * (-35°C) = -10357 J

For the cold water:

m(cold) = 70 g
c(water) = 4.18 J/g°C (specific heat capacity of water)
ΔT(cold) = 45°C - 20°C = 25°C

q(cold) = 70 g * 4.18 J/g°C * 25°C = 7295 J

Now, the heat gained or lost by the calorimeter, q(calorimeter), is the negative of the heat gained or lost by the water:

q(calorimeter) = -q(water) = -(-10357 J + 7295 J) = 3062 J

Finally, we can calculate the heat capacity of the calorimeter using the formula:

C(calorimeter) = q(calorimeter) / ΔT(calorimeter)

where C(calorimeter) is the heat capacity of the calorimeter and ΔT(calorimeter) is the change in temperature of the calorimeter.

ΔT(calorimeter) = T(final) - T(initial) = 45°C - 25°C = 20°C

C(calorimeter) = 3062 J / 20°C = 153.1 J/°C

Therefore, the heat capacity of the calorimeter is 153.1 J/°C.