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 capacity is ?J/degree C.
heat gain by cool water + heat lost by warm wataer + heat gained by calorimeter = 0
[mass cool water x specific heat x (Tfinal-Tinitial)] + [mass warm water x specific heat x (Tfinal-Tinitial)] + [Ccal*(Tfinal-Tinitial)] = 0
To determine the heat capacity of the calorimeter, we need to consider the heat gained or lost by each portion of water and the heat exchanged between them.
First, we calculate the heat gained or lost by the 50 g of water at 20°C. We use the specific heat capacity of water, which is approximately 4.18 J/g°C.
q1 = m1 * c * ΔT1
= 50 g * 4.18 J/g°C * (45°C - 20°C)
= 4350 J
Next, we calculate the heat gained or lost by the 50 g of water at 80°C.
q2 = m2 * c * ΔT2
= 50 g * 4.18 J/g°C * (45°C - 80°C)
= -4350 J
Note that the negative sign indicates heat loss.
Lastly, we need to calculate the heat exchanged between the portions of water, which is equal in magnitude but opposite in sign to the heat lost by one portion and gained by the other.
q exchanged = -q2
= -(-4350 J)
= 4350 J
The heat exchanged is also given by the heat capacity of the calorimeter (Ccal) multiplied by the temperature change of the mixed portions of water (ΔTmixed).
q exchanged = Ccal * ΔTmixed
Rearranging the equation, we can solve for Ccal:
Ccal = q exchanged / ΔTmixed
= 4350 J / (45°C - 45°C)
= 4350 J / 0°C
= undefined
The heat capacity of the calorimeter cannot be determined based on the given information. It is likely that there may be additional information or steps in the experiment's procedure that would allow us to calculate the value.