Heat of Reaction Experiment 23

• Purpose: To observe and measure heat transfer between liquids from a chemical reaction.

• Calorimetry – The measurement of heat effects in chemical and physical processes.

Specific Heat

• Specific Heat– The amount of heat required to raise 1 gram of substance 1 C.

• Note: Specific heat of water = 1 calorie /(gram C)

• The device used to measure the transfer of heat is the calorimeter.

Three types of Heat

• Conduction – When two bodies are next to each other. No mixing of liquids or material. Two cans of soda sitting next to each other (one is hot and one is cold).

• Convection – an exchange of molecules; both cans of soda are poured into one large glass. The colder liquid absorbs heat while the warmer liquid releases it. Also a pot of boiling water.

• Radiation – light from the sun is absorbed and heats an object. A sandy beach in the summer.

Q = MC(T)

• Q = heat flow for hot (QH) or cold (QC) liquid

• M = mass of water (grams)

• C = specific heat of water (1 cal/g C)T = (Tf –Ti) (in C)

Qh = 50g 1 cal/g C (32 C – 40 C) = - 400 calories “Losing heat”

Qc = 50g 1 cal/g C (32 C – 25 C) = 350 cal “Gaining heat”

Compute Qhot and Qcold

Example # 1 : 50g of hot water at 40 C is mixed with 50 g of cool water at 25 C. After mixing, the final temperature was 32 C.

Q = MC(T) T= Tf-Ti

Qhot does not equal Qcold ? why

Heat was absorbed by your calorimeter?

From the last problem:Qh = - 400 calories Qc = 350 cal

Q cal = -[Qh + Qc]Q cal = -[- 400 cal + 350 cal] = +50 calories Q cal = +50 calories stolen by calorimeter

•Each calorimeter absorbs a different amount of heat.Use the same calorimeter for part #2.

Use Q cal calories stolen by calorimeter to determine Hc (Heat Capacity) of calorimeter

From the last problem:Q cal = +50 calories

We must calculate Hc (Heat Capacity): this will be a correction factor for part #2.

•Hc = Q cal

T for the cold water.

From my trial: Hc = 50 cal / 7.0 OC = 7.1cal/ OC

Part One:• Determine the amount of heat that your calorimeter absorbs.• Then calculate Calculate Hc (Heat Capacity) of calorimeter

• HCl + NaOH NaCl + H2O + Heat Energy

(Higher in energy) (Lower in energy) • Note: This reaction is exothermic thus energy is

released.

Determine the heat released in a chemical reaction when 20.0g of HCl at 22 C was added to 20.0g of NaOH at 22 C. The final temperature was 28 C.Same calorimeter as in Part I; use Hc from Part I.7.1cal/ OC

Part Two:•Determine the amount of energy released during a chemical reaction.

QRx = -[Q Observed+ HcΔT]

Q Observed = what we recorded during the reaction. Q Rx = what is actually released during the reaction

So we have: Q Rx = -[McΔT + HcΔT] watercold Δtemp.of

rcalorimeteby absorbed cal.Hc

282cal676 0.140 Ocal

COg

cal CCgQ O

C

ORx

HC ΔTΔT

QObserved=McΔT

We can determine the amount of heat absorbed by the environment by using: QRx = - QSurr

Q Observed = what we recorded during the reaction. Qsurr =what is absorbed in the environment Q Rx = what is released by the reaction

-282cal = - Q Surr

If: QRx=-282 cal

SurrQ282cal

282cal1

282cal

Today in Lab

• Part # 1 • Determine Hc for your calorimeter • Add 80 ml of water @ room temp to your

cal. • Heat water 80g of water to 60OC • Once at temperature remove from heat. • Record initial temps for both the hot and

cold water. In tables 23-2 & 3

Once at 60OC.

• Start the timer, and record your initial temperature in table 23-1.

• Continue to record the temp of your hot water every 30 seconds for 2.5 minutes.

• At 150 sec (2.5 min) pour the hot water into the calorimeter and continue to record the temperatures.

• Do two trials, and use the better of the two for part #2.

• Graph the data that you use in part #2.

Part # 2 Determine the amount of heat that is generated from the reaction.

• Add 50mL. of HCl into the calorimeter.

• Pour 50mL. of NaOH into a beaker.

• Record the initial temps. of both and record data in table 23-5

• Pour the NaOH into the calorimeter, turn on the timer, mix, and record temps.

Time (s) Temp. OC

0 49

30 48

60 47

90 46

120 45

150 Mix

180 30

21 29

240 28

270 27

23-1

temp vs time

0

10

20

30

40

50

60

time (s)

tem

p (

c)

Series1

Series2

Series1 49 48 47 46 45 30 29

Series2 25 25 25 25 25 30 29

0 30 60 90 120 150 180 210

4425

31

trial Ti Tf ΔT

1 49OC 31OC -18OC

trial Ti Tf ΔT

1 25OC 31OC 6OC

23-2

23-3

23-4

Time (s) Temp. OC

0 20

60 27

120 28

180 28

240 28

300 28

Trial T i Tf Δ T

1 20 0C 28 0C +8 0C

23-5

Due next week

• Your calculations for each trial in part #1, data and calculations from part # 2.

• Show all work for full credit. • Two graphs one for each part. • A graphs done on graph paper will be accepted this

time.