Decrease $2$ degrees Celsius

Decrease $5$ degrees celsius

$0$

$518$

$26\mathrm{.}60$

$764$

$52\mathrm{.}82$

$1049$

$90$

$1430$

$157$

$1953$

$244$

$2615$

$308$

$3435$

$5322$

Calculations

You will treat temperature as if it were a concentration $($of heat$).$ To analyze the data, you will need to know the difference between each temperature value recorded and the original room temperature. For each temperature value recorded, determine the difference between the temperature of the hot water and the room temperature.

Record this difference in your data table.

Temperature difference $(-2)$

Temperature difference $(-5)$

$80-29=51$

$63-29=34$

$78-29=49$

$58-29=29$

$76-29=47$

$53-29=24$

$74-29=45$

$48-29=19$

$72-29=43$

$43-29=14$

$70-29=41$

$38-29=9$

$68-29=39$

$33-29=4$

$28-29=-1$

A downloadable version of Vernier's Logger Pro $3$ graphing software can be used to plot three graphs using this temperature difference versus time data $($you may use any graphing program such as MS Excel or Google Sheets$-$ be sure to paste them onto a Word Document or PDF$).$

$($Directions for downloading and using Logger Pro $3)$

You can also use another graphing tool, such as those found in Excel or PowerPoint, to graph the data.

Create the $3$ different graphs listed below $($be sure to consult the graph rubric for the course in order to determine the specifics for graph construction$)$ and paste all graphs you construct into a Word document or pdf to include in the lab report.

Temperature difference vs time

ln Temperature difference vs$.$ time $($natural log of the temperature difference versus time$)$

Time $($seconds$)$

ln$($T$)$

$0$

$0$

$26\mathrm{.}60$

$3\mathrm{.}281$

$52\mathrm{.}82$

$3\mathrm{.}967$

$90$

$4\mathrm{.}4998$

$157$

$5\mathrm{.}056$

$244$

$5\mathrm{.}497$

$308$

$5\mathrm{.}730$

$518$

$6\mathrm{.}2499$

$764$

$6\mathrm{.}639$

$1049$

$6\mathrm{.}9556$

$1430$

$7\mathrm{.}265$

$1953$

$7\mathrm{.}577$

$2615$

$7\mathrm{.}869$

$3435$

$8\mathrm{.}142$

$5322$

$8\mathrm{.}5796$

Temperature difference$-1$ vs$.$ time $($the reciprocal of the temperature difference versus time$)$

Time $($seconds$)$

$1/$T

$0$

$1/0=0$

$26\mathrm{.}60$

$1/26\mathrm{.}60=0\mathrm{.}0376$

$52\mathrm{.}82$

$1/52\mathrm{.}82=0\mathrm{.}0189$

$90$

$1/90=0\mathrm{.}0111$

$157$

$1/157=0\mathrm{.}0064$

$244$

$1/244=0\mathrm{.}0041$

$308$

$1/308=0\mathrm{.}00325$

$518$

$1/518=0\mathrm{.}0019$

$764$

$1/764=0\mathrm{.}00131$

$1049$

$1/1049=0\mathrm{.}0009539$

$1430$

$1/1430=0\mathrm{.}0006993$

$1953$

$1/1953=0\mathrm{.}0005120$

$2615$

$1/2615=0\mathrm{.}0003824$

$3435$

$1/3435=0\mathrm{.}0002911$

$5322$

$1/5322=0\mathrm{.}0001879$

Based on the information above, answer these questions:

Determine the order of the reaction from the graphs' construction in the lab. Explain.

Using the graph that determined the order of the reaction, calculate the slope of the line; this is your value for k$,$ the rate constant. Include the units for k$.$

Write the rate law for the cooling of water. This should take the form of with the values for k and n substituted.

Using what you know about heat $($review Chapter $5,$ if needed$)$ and kinetic energy, and what you have learned from Lesson $2$ on order of reactions, justify how the order of the reaction $($determined above$)$ matches the changes in kinetic energy.

The above procedure was repeated by a student whose AC broke on a hot summer day. Would the rate constant increase, decrease or remain the same as the observed results? Explain.

Explain why the volume of water would not change the calculated rate constant.