Assume you are performing the calibration step of Experiment 8 and you begin with 50 g of water at 20 oC and 50 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?

heat gained by cold water + heat gained by calorimeter = heat lost by hot water.

Calculate heat lost by hot water.
q = mass x specific heat x (Tfinal-Tinitial)

Calculate heat gained by cold water.
q = mass x specific heat x (Tfinal-Tinitial)

The difference is the heat gained by the calorimeter. heat capacity = delta H/delta T.

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To determine the heat capacity of the calorimeter, you need to use the principle of energy conservation. The heat gained by the cooler water is equal to the heat lost by the hotter water. This can be expressed as:

Heat gained by cooler water = Heat lost by hotter water

The heat gained by the cooler water can be calculated using the equation:

q(cooler) = m(cooler) * C(cooler) * ΔT(cooler)

Where:
q(cooler) is the heat gained by the cooler water,
m(cooler) is the mass of the cooler water,
C(cooler) is the specific heat capacity of water, which is approximately 4.18 J/g°C,
ΔT(cooler) is the change in temperature of the cooler water.

The heat lost by the hotter water can be calculated using the equation:

q(hotter) = m(hotter) * C(hotter) * ΔT(hotter)

Where:
q(hotter) is the heat lost by the hotter water,
m(hotter) is the mass of the hotter water,
C(hotter) is the specific heat capacity of water,
ΔT(hotter) is the change in temperature of the hotter water.

Since the heat gained and lost are equal, you can set up the equation as follows:

m(cooler) * C(cooler) * ΔT(cooler) = m(hotter) * C(hotter) * ΔT(hotter)

We know the following information:

m(cooler) = 50 g
ΔT(cooler) = (final temperature of mixed water - initial temperature of cooler water)

m(hotter) = 50 g
ΔT(hotter) = (final temperature of mixed water - initial temperature of hotter water)

Plugging in the given values:

ΔT(cooler) = (45 oC - 20 oC) = 25 oC
ΔT(hotter) = (45 oC - 80 oC) = -35 oC (the temperature change is negative as the hotter water is getting cooler)

Now we can rearrange the equation and solve for the specific heat capacity of the calorimeter:

C(calorimeter) = (m(cooler) * C(cooler) * ΔT(cooler)) / (m(hotter) * ΔT(hotter))

Plugging in the given values:

C(calorimeter) = (50 g * 4.18 J/g°C * 25 oC) / (50 g * -35 oC)

Simplifying further:

C(calorimeter) = (2090 J°C) / (-1750 g°C)

Calculating the final result:

C(calorimeter) ≈ -1.19 J/°C

Note: The negative sign indicates that the calorimeter loses heat as it absorbs heat from the water.