When solutions of two reactants were mixed in a coffee-cup calorimeter, the following temperatures were recorded as a function of time. Plot the data on one of the available pieces of graph paper. Obtain tf, the final temperature, by extrapolating to the time of mixing (time = 0 s) with a straight line. The initial temperature, ti, was 24.3°C.
Time (s) t(°C) Time (s) t(°C)
30 38.6 150 39.3
60 39.7 180 39.2
90 39.5 210 39.1
120 39.4 240 39.0
Would I use this equation:
ΔT = Tfinal - Tinitial?
rgbgt
24.6
34
45
To plot the data on a graph, you can use a scatter plot. Label the x-axis with "Time (s)" and the y-axis with "Temperature (°C)".
Next, plot the data points on the graph using the given time-temperature pairs:
(30, 38.6), (60, 39.7), (90, 39.5), (120, 39.4), (150, 39.3), (180, 39.2), (210, 39.1), (240, 39.0)
Connect the data points with line segments to create a smooth curve.
To extrapolate the final temperature (tf) at the time of mixing (time = 0s), we can extend a straight line from the last two data points (240s, 39.0°C) and (210s, 39.1°C) back to the y-axis (time = 0s).
To do this, find the slope of the line using the formula:
slope = (change in temperature) / (change in time)
slope = (39.1 - 39.0) / (210 - 240) = -0.05°C/s
With the slope known, we can use the point-slope form of a line to find the equation of the straight line:
y - y1 = m(x - x1)
y = -0.05x + b
Using the point (210, 39.1), we can substitute the values into the equation:
39.1 = -0.05(210) + b
Solve for b:
b = 39.1 + 0.05(210) = 39.1 + 10.5 = 49.6
Now we have the equation for the straight line:
y = -0.05x + 49.6
To find the final temperature (tf) at time = 0s, substitute x = 0 into the equation:
tf = -0.05(0) + 49.6 = 49.6
Therefore, the final temperature (tf) at the time of mixing is 49.6°C.