Question

temperature change & intermolecular forces (simulated \specific heat\ comparison)
instead of hot plates/thermometers, use the simulated heating data below. both liquids receive the same energy input for the same time.
heating results (quantitative)

1. compute $\delta t$ for each liquid and write your answer in the data set below.

liquid	initial temp (°c)	final temp after equal heating (°c)	$\delta t$ (°c)
paraffin	22.0	41.0

properties of polar vs nonpolar compounds lab (lab report)

Explanation:

Step1: Calculate ΔT for Water

The formula for temperature change is \( \Delta T = \text{Final Temp} - \text{Initial Temp} \). For water, initial temp is \( 22.0^\circ\text{C} \) and final temp is \( 30.0^\circ\text{C} \). So, \( \Delta T = 30.0 - 22.0 = 8.0 \).

Step2: Calculate ΔT for Paraffin

Using the same formula, for paraffin, initial temp is \( 22.0^\circ\text{C} \) and final temp is \( 41.0^\circ\text{C} \). So, \( \Delta T = 41.0 - 22.0 = 19.0 \).

Answer:

For Water, \( \Delta T = 8.0^\circ\text{C} \); For Paraffin, \( \Delta T = 19.0^\circ\text{C} \)

Filling the table:

Liquid	Initial Temp (\(^\circ\text{C}\))	Final Temp after equal heating (\(^\circ\text{C}\))	\( \Delta T \) (\(^\circ\text{C}\))
Paraffin	22.0	41.0	19.0