I need to find the calorimeter constant for this experiment.
Mass of nested cups (two cups inside one another): 3.04 g
Mass of single cups: 1.51 g
75 mL of cold water goes into nested cup: 76.045 g and the temp. remained constant at 23.0
75 ml of warm water into single cup: 72.755 g and the temp. dropped from 64.75 to 61.25
75 mL of warm water and 75 mL of cold water are placed into the nested cups and a cup is placed overtop of the nested cups to create the calorimeter. The change in temp. for the hot and cold water mixed was 38.5 to 40.75
To find the calorimeter constant for this experiment, you can use the principle of energy conservation. The heat lost by the warm water is equal to the heat gained by the cold water and the calorimeter.
First, calculate the heat lost by the warm water. You can use the equation:
Q = mcΔT
Where Q is the heat lost, m is the mass of the warm water, c is the specific heat capacity of water, and ΔT is the change in temperature.
In this case, the mass of the warm water is 75 g (75 ml of water is assumed to have the same mass), the specific heat capacity of water is 4.18 J/g°C, and the change in temperature is 3.5°C (from 64.75 to 61.25).
Q_warm = 75 g * 4.18 J/g°C * 3.5°C = 904.5 J
Next, calculate the heat gained by the cold water. Again, you can use the same equation:
Q_cold = mcΔT
In this case, the mass of the cold water is also 75 g, the specific heat capacity of water is 4.18 J/g°C, and the change in temperature is 17.75°C (from 23.0 to 40.75).
Q_cold = 75 g * 4.18 J/g°C * 17.75°C = 5596.63 J
Since energy is conserved, the heat gained by the cold water and the calorimeter is equal to the heat lost by the warm water:
Q_cold + Q_calorimeter = Q_warm
To find the calorimeter constant, we need to rearrange the equation:
Q_calorimeter = Q_warm - Q_cold
Q_calorimeter = 904.5 J - 5596.63 J
Q_calorimeter = -4692.13 J
The negative sign indicates that the calorimeter absorbed the heat rather than releasing it. Therefore, the calorimeter constant for this experiment is -4692.13 J.