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?

whats the answer

I don't know what experiment 8 is nor do I know the procedure. However, if I know the temperature of the calorimeter initially I can help you. Perhaps the experiment started with 70 g H2O at 20 C. Let me make that assumption and you can correct me if I'm wrong.

[mass cool H2O x specific heat x (Tf-Ti)] + [mass warm H2O x specific heat x (Tf-Ti)] + Ccal(Tf-Ti) = 0
Substitute 70 for mass cool and mass warm water, specific heat, and Tf and Ti. Solve for Ccal.

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

Q = mcΔT

where Q represents the heat transfer, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.

In this case, we have two portions of water with masses of 70 g each, so the total mass of water is 70 g + 70 g = 140 g.

We can calculate the heat absorbed by the water:

Q_water = mcΔT = (140 g)(4.18 J/g·°C)(45 °C - 45 °C) = 0 J

We know that the heat absorbed by the water is equal to the heat released by the calorimeter:

Q_calorimeter = -Q_water = 0 J

Since the heat capacity of the calorimeter is defined as the amount of heat required to raise its temperature by 1 degree Celsius, we can solve for the heat capacity:

C_calorimeter = Q_calorimeter / ΔT_calorimeter

However, in this case, the temperature of the mixed portions of water did not change, so the change in temperature ΔT_calorimeter is 0.

Therefore, C_calorimeter = Q_calorimeter / ΔT_calorimeter = 0 J / 0 °C = indeterminate.

Since the temperature did not change, we cannot determine the heat capacity of the calorimeter based on this information alone.