For each trial, enter the amount of heat lost by the calorimeter, qcalorimeter.
Be careful of the algebraic sign here and remember that the change in temperature is equal to the final temperature minus the initial temperature.
Report your answer using 4 digits. Note, this is 1 or 2 digits beyond the correct number of significant figures.
Trial # Initial Masswater Tiwater Tf qwater qcalorimeter
#1: 86.281 51.2 23.7 -9927.
#2: 87.777 49.0 29.0 -7345.
#3: 89.997 52.2 27.2 -9413.
#1: -9927.
#2: -7345.
#3: -9413.
To calculate the amount of heat lost by the calorimeter (qcalorimeter), you need to use the equation:
qcalorimeter = -(qwater)
Where qwater is the amount of heat gained or lost by the water, which can be calculated using the equation:
qwater = masswater * specific heatwater * ΔTwater
Here, masswater is the initial mass of water, specific heatwater is the specific heat capacity of water, and ΔTwater is the change in temperature of the water.
Given the data provided, you can calculate qwater for each trial as follows:
Trial #1:
masswater = 86.281 g
specific heatwater = assumed value (you need to provide this value)
ΔTwater = Tf - Tiwater = 23.7 °C - 51.2 °C = -27.5 °C (be careful with the algebraic sign)
Substituting these values into the equation, you can find qwater for trial #1.
Similarly, you can calculate qwater for trial #2 and trial #3 using the given data.
Once you find the values of qwater, you can use the equation qcalorimeter = -(qwater) to get the values of qcalorimeter for each trial.
To find the amount of heat lost by the calorimeter (qcalorimeter), we need to use the formula:
qcalorimeter = qwater + qsurroundings
However, since qsurroundings is not given in the question, we can assume that it is negligible and can ignore it. Therefore, we can simplify the formula to:
qcalorimeter = qwater
Using this formula, we can calculate qcalorimeter for each trial:
In Trial #1:
qcalorimeter = -9927.0000 J
In Trial #2:
qcalorimeter = -7345.0000 J
In Trial #3:
qcalorimeter = -9413.0000 J