Download presentation

1
Calorimetry

2
**Heat Transfer Surroundings Final Temperature Block “A” Block “B”**

SYSTEM 20 g (40oC) 20 g (20oC) 30oC Al Al m = 20 g T = 40oC m = 20 g T = 20oC What will be the final temperature of the system ? a) 60oC b) 30oC c) 20oC d) ? Assume NO heat energy is “lost” to the surroundings from the system.

3
**Heat Transfer ? Surroundings Final Temperature Block “A” Block “B”**

SYSTEM 20 g (40oC) 20 g (20oC) 30.0oC 20 g (40oC) 10 g (20oC) 33.3oC Al Al m = 20 g T = 40oC m = 10 g T = 20oC What will be the final temperature of the system ? a) 60oC b) 30oC c) 20oC d) ? Assume NO heat energy is “lost” to the surroundings from the system.

4
**Heat Transfer Surroundings Final Temperature Block “A” Block “B”**

SYSTEM 20 g (40oC) 20 g (20oC) 30.0oC 20 g (40oC) 10 g (20oC) 33.3oC 20 g (20oC) 10 g (40oC) 26.7oC Al Al m = 20 g T = 20oC m = 10 g T = 40oC Assume NO heat energy is “lost” to the surroundings from the system.

5
**Heat Transfer Surroundings Final Temperature Block “A” Block “B”**

SYSTEM 20 g (40oC) 20 g (20oC) 30.0oC 20 g (40oC) 10 g (20oC) 33.3oC 20 g (20oC) 10 g (40oC) 26.7oC H2O Ag m = 75 g T = 25oC m = 30 g T = 100oC Real Final Temperature = 26.6oC Why? We’ve been assuming ALL materials transfer heat equally well.

6
**Specific Heat Water and silver do not transfer heat equally well.**

Water has a specific heat Cp = J/goC Silver has a specific heat Cp = J/goC What does that mean? It requires Joules of energy to heat 1 gram of water 1oC and only Joules of energy to heat 1 gram of silver 1oC. Law of Conservation of Energy… In our situation (silver is “hot” and water is “cold”)… this means water heats up slowly and requires a lot of energy whereas silver will cool off quickly and not release much energy. Lets look at the math!

7
**Calorimetry “loses” heat Surroundings SYSTEM Tfinal = 26.6oC H2O Ag**

m = 75 g T = 25oC m = 30 g T = 100oC

8
**Calorimetry Surroundings SYSTEM H2O Ag m = 75 g T = 25oC m = 30 g**

9
**Find the mass of the iron.**

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC). When thermal equilibrium is reached, the system has a temperature of 42oC. Find the mass of the iron. Fe T = 500oC mass = ? grams T = 20oC mass = 240 g - LOSE heat = GAIN heat - [(Cp,Fe) (mass) (DT)] = (Cp,H2O) (mass) (DT) - [( J/goC) (X g) (42oC - 500oC)] = (4.184 J/goC) (240 g) (42oC - 20oC)] Drop Units: - [(0.4495) (X) (-458)] = (4.184) (240 g) (22) 205.9 X = X = g Fe Calorimetry Problems 2 question #5

10
**A 97 g sample of gold at 785oC is dropped into 323 g of water, which has an initial**

temperature of 15oC. If gold has a specific heat of J/goC, what is the final temperature of the mixture? Assume that the gold experiences no change in state of matter. Au T = 785oC mass = 97 g T = 15oC mass = 323 g LOSE heat = GAIN heat - - [(Cp,Au) (mass) (DT)] = (Cp,H2O) (mass) (DT) Drop Units: - [(0.129 J/goC) (97 g) (Tf - 785oC)] = (4.184 J/goC) (323 g) (Tf - 15oC)] - [(12.5) (Tf - 785oC)] = (1.35x 103) (Tf - 15oC)] -12.5 Tf x = x 103 Tf x 104 3 x 104 = x 103 Tf Tf = oC Calorimetry Problems 2 question #8

11
**If 59 g of water at 13oC are mixed with 87 g of water at 72oC, find the final temperature**

of the system. T = 13oC mass = 59 g T = 72oC mass = 87 g LOSE heat = GAIN heat - - [(Cp,H2O) (mass) (DT)] = (Cp,H2O) (mass) (DT) Drop Units: - [(4.184 J/goC) (59 g) (Tf - 13oC)] = (4.184 J/goC) (87 g) (Tf - 72oC)] - [(246.8) (Tf - 13oC)] = (364.0) (Tf - 72oC)] Tf = Tf = Tf Tf = oC Calorimetry Problems 2 question #9

12
**A 322 g sample of lead (specific heat = 0**

A 322 g sample of lead (specific heat = J/goC) is placed into 264 g of water at 25oC. If the system's final temperature is 46oC, what was the initial temperature of the lead? T = ? oC Pb mass = 322 g Ti = 25oC mass = 264 g Pb Tf = 46oC - LOSE heat = GAIN heat - [(Cp,Pb) (mass) (DT)] = (Cp,H2O) (mass) (DT) Drop Units: - [(0.138 J/goC) (322 g) (46oC - Ti)] = (4.184 J/goC) (264 g) (46oC- 25oC)] - [(44.44) (46oC - Ti)] = (1104.6) (21oC)] Ti = 44.44 Ti = Ti = 568oC Calorimetry Problems 2 question #12

13
**Calorimetry with Change in State**

14
**A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC. **

Find the system's final temperature. Temperature (oC) 40 20 -20 -40 -60 -80 -100 120 100 80 60 140 Time DH = mol x DHfus DH = mol x DHvap Heat = mass x Dt x Cp, liquid Heat = mass x Dt x Cp, gas Heat = mass x Dt x Cp, solid ice T = -11oC mass = 38 g D water cools T = 56oC mass = 214 g B warm water A C warm ice melt ice LOSE heat = GAIN heat - - [(Cp,H2O) (mass) (DT)] = (Cp,H2O) (mass) (DT) + (Cf) (mass) + (Cp,H2O) (mass) (DT) - [(4.184 J/goC)(214 g)(Tf - 56oC)] = (2.077 J/goC)(38 g)(11oC) + (333 J/g)(38 g) + (4.184 J/goC)(38 g)(Tf - 0oC) - [(895) (Tf - 56oC)] = (159) (Tf)] - 895 Tf = Tf - 895 Tf = Tf = Tf Tf = oC Calorimetry Problems 2 question #10

15
**qD = (4.184 J/goC) (238.4 g) (Tf - 8oC)**

(1000 g = 1 kg) 25 g of 116oC steam are bubbled into kg of water at 8oC. Find the final temperature of the system. 238.4 g - [qA + qB + qC] = qD - [(Cp,H2O) (mass) (DT)] (Cv,H2O) (mass) + (Cp,H2O) (mass) (DT) = [(Cp,H2O) (mass) (DT)] qD = (4.184 J/goC) (238.4 g) (Tf - 8oC) qD = Tf qA = [(Cp,H2O) (mass) (DT)] qB = (Cv,H2O) (mass) qC = [(Cp,H2O) (mass) (DT)] qA = [(2.042 J/goC) (25 g) (100o - 116oC)] qA = (2256 J/g) (25 g) qC = [(4.184 J/goC) (25 g) (Tf - 100oC)] qA = J qA = J qA = 104.5Tf - [qA + qB + qC] = qD - [ Tf ] = 997Tf Temperature (oC) 40 20 -20 -40 -60 -80 -100 120 100 80 60 140 Time DH = mol x DHfus DH = mol x DHvap Heat = mass x Dt x Cp, liquid Heat = mass x Dt x Cp, gas Heat = mass x Dt x Cp, solid Tf = 997Tf Tf = 997Tf A = Tf C 1102 B Tf = 68.6oC D Calorimetry Problems 2 question #11

16
**of the system is 24oC, what was the mass of the ice?**

A sample of ice at –12oC is placed into 68 g of water at 85oC. If the final temperature of the system is 24oC, what was the mass of the ice? T = -12oC H2O mass = ? g Ti = 85oC mass = 68 g ice Tf = 24oC GAIN heat = - LOSE heat qA = [(Cp,H2O) (mass) (DT)] qA = [(2.077 J/goC) (mass) (12oC)] 24.9 m [ qA + qB + qC ] = - [(Cp,H2O) (mass) (DT)] [ qA + qB + qC ] = - [(4.184 J/goC) (68 g) (-61oC)] qB = (Cf,H2O) (mass) qB = (333 J/g) (mass) 333 m 458.2 m = 458.2 qC = [(Cp,H2O) (mass) (DT)] qC = [(4.184 J/goC) (mass) (24oC)] 100.3 m m = g qTotal = qA + qB + qC 458.2 m Calorimetry Problems 2 question #13

17
**A Coffee Cup Calorimeter**

Thermometer Glass stirrer Cork stopper Two Styrofoam ® cups nested together containing reactants in solution Thermometer Styrofoam cover cups Stirrer A Coffee Cup Calorimeter Thermal energy cannot be measured easily. Temperature change caused by the flow of thermal energy between objects or substances can be measured. Calorimetry is the set of techniques employed to measure enthalpy changes in chemical processes using calorimeters. Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 302

18
**Bomb Calorimeter thermometer stirrer full of water ignition wire**

Constant — volume calorimetry – To study reactions in which one or more of the reactants is a gas, such as a combustion reaction, a constant-volume calorimeter is used to measure the enthalpy changes that accompany these reactions. – Bomb calorimeter operates at constant volume and measures enthalpies of combustion. – Heat released by a reaction carried out at constant volume is identical to the change in internal energy (ΔE) rather than the enthalpy change (ΔH). – Assuming ΔE = ΔH, the relationship between the measured temperature change and ΔHcomb is ΔHcomb = qcomb = –qcalorimeter = – CbombΔT. steel “bomb” sample

19
A Bomb Calorimeter

Similar presentations

OK

1. During the phase change from solid to liquid, the temperature of the sample a)increases b)decreases c)doubles d)halves e)does not change.

1. During the phase change from solid to liquid, the temperature of the sample a)increases b)decreases c)doubles d)halves e)does not change.

© 2018 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google