Question

in part 1 of this lab, you will be calculating δrh by conducting constant - pressure calorimetry measurements.
one of the steps you will take is calculating δt of the water inside the test tube and then using the formula qcal = m x cs x δt.
what is the correct procedure for calculating δt?
select one:
○ leave tinitial and tfinal in °c, then calculate tfinal - tinitial, then leave the result in °c (as we learned in lecture)
○ leave tinitial and tfinal in °c, then calculate tfinal - tinitial in °c, then add 273 to convert the result to k (please dont do this)

Explanation:

Step1: Understand temperature change formula

The formula for temperature change is \(\Delta T=T_{final}-T_{initial}\). When dealing with temperature differences in calorimetry (constant - pressure in this case), the unit of temperature difference is the same as the unit of the initial and final temperatures. Since \(T_{initial}\) and \(T_{final}\) are given in \(^{\circ}C\), the difference \(\Delta T\) is also in \(^{\circ}C\).

Step2: Analyze the second option

Adding 273 to convert the temperature difference (which is already in \(^{\circ}C\)) to Kelvin is incorrect. The conversion formula from Celsius to Kelvin is \(T(K)=T(^{\circ}C)+ 273\), but this is for absolute temperatures, not for temperature differences. For example, if \(T_1 = 10^{\circ}C=(10 + 273)K=283K\) and \(T_2=20^{\circ}C=(20 + 273)K = 293K\), then \(\Delta T\) in \(^{\circ}C\) is \(20 - 10=10^{\circ}C\) and in \(K\) is \(293 - 283=10K\). So, we don't need to add 273 for the temperature difference.

Answer:

Leave \(T_{initial}\) and \(T_{final}\) in \(^{\circ}C\), then calculate \(T_{final}-T_{initial}\), then leave the result in \(^{\circ}C\) (as we learned in lecture)