Assume you are performing the calibration step of Experiment 8 and you begin with 20 g of water at 20 oC and 20 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, you need to use the principle of energy conservation. In this case, you can assume that no heat is lost to the surroundings during the experiment.
The equation you will use is:
(m1 * c1 * ΔT1) + (m2 * c2 * ΔT2) = (m_cal * c_cal * ΔT_cal)
Where:
m1 is the mass of water at 20 °C
c1 is the specific heat capacity of water at constant pressure (4.18 J/g°C)
ΔT1 is the change in temperature for the water at 20 °C (final temperature - initial temperature)
m2 is the mass of water at 80 °C
c2 is the specific heat capacity of water at constant pressure (4.18 J/g°C)
ΔT2 is the change in temperature for the water at 80 °C (final temperature - initial temperature)
m_cal is the mass of the calorimeter (unknown)
c_cal is the specific heat capacity of the calorimeter (unknown)
ΔT_cal is the change in temperature of the calorimeter (final temperature - initial temperature)
Given:
m1 = 20 g
c1 = 4.18 J/g°C
ΔT1 = 45 °C - 20 °C = 25 °C
m2 = 20 g
c2 = 4.18 J/g°C
ΔT2 = 45 °C - 80 °C = -35 °C (negative because the water is cooling down)
ΔT_cal = 45 °C - 25 °C = 20 °C
Using the equation, you can solve for m_cal * c_cal:
(20 g * 4.18 J/g°C * 25 °C) + (20 g * 4.18 J/g°C * -35 °C) = (m_cal * c_cal * 20 °C)
(2085 J) + (-2918 J) = (20 °C * m_cal * c_cal)
-833 J = (20 °C * m_cal * c_cal)
Now, you need to convert the units to joules:
-833 J = (20 °C * m_cal * c_cal * 4.18 J/g°C)
Divide both sides of the equation by 20 °C * 4.18 J/g°C:
-833 J / (20 °C * 4.18 J/g°C) = m_cal * c_cal
Solve for m_cal * c_cal:
m_cal * c_cal = -9.99 g°C
Therefore, the heat capacity of the calorimeter is approximately -9.99 J/g°C.