Designing a Handwarmer Lab Report (AP Chemistry) Essay

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Lab 5

Julia (hour 2)

TITLE: INVESTIGATION – Hand Warmer Design Challenge: Where does the heat come from?
PURPOSE: to determine which of the 3 ionic compounds (NaCl, LiCl, or NaCH3COO) is most suitable for use as a hand warmer

PROCEDURE:
DAY 1 (Part 2 only):

1) Measure out 2 separate samples of 100.0 mL of distilled water

2) Heat one to about 50˚C, and place other one in calorimeter (at around 20˚C)

3) Add heater water to calorimeter, cover top, wait 15 seconds, measure temp

4) Repeat

DAY 2:
1) measure 40mL of distilled water into a styrofoam cup on stir plate
2) record temp of water
3) measure out approximately 4g of salt
4) add to water and stir
5)
…show more content…
The hot water is losing energy and the enthalpy change is negative.

1c) q-cal

Average q-cold:
Average q-hot:

qcold = 11800 J qhot = -11900 J

Lab 5!

Julia (hour 2)

qhot = - ( qcold + qcal )
(-11900 J) = - (11800 J + qcal )
11900 J = 11800 J + qcal

1d) C

2) qsol/qrxn

100. J = qcal
Trial 1:
∆Tcal = (35.7˚C - 21.6˚C)

= 14.1˚C
Trial 2:
∆Tcal = (35.8˚C - 21.6˚C)

= 14.2˚C
Average:

∆Tcal = (14.1˚C) + (14.2˚C) = 14.2˚C

2
C

= qcal / ∆Tcal
= (100. J) / (14.2˚C)
= 2.60 J/˚C
NaCl

Trial 1:

qsol

= ( m ) ( c ) ( ∆T )

= (40.0 g + 4.01 g)(4.184 J/g˚C)(21.1˚C - 22.6˚C)

= -280 J

qrxn

= -qsol - C∆T

= -(-280 J) - (2.60 J/˚C)(21.1˚C - 22.6˚C)

= 280 J

Trial 2:

qsol

= ( m ) ( c ) ( ∆T )

= (40.0 g + 4.05 g)(4.184

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