q = m Δt C p q = (40.0 g) (20.0 °C) (4.184 J g¯ 1 °C¯ 1 ) q = 3347.2 J

q = m Δt C p q = (40.0 g) (15.0 °C) (4.184 J g¯ 1 °C¯ 1 ) q = 2510.4 J

3347.2 − 2510.4 = 836.8 J

836.8 J / 15.0 °C = 55.8 J / °C

q = m Δt C p q = (25.0 g) (25.0 °C) (4.184 J g¯ 1 °C¯ 1 ) q = 2615.0 J

q = m Δt C p q = (25.0 g) (10.0 °C) (4.184 J g¯ 1 °C¯ 1 ) q = 1046.0 J

2615.0 − 1046.0 = 1569.0 J

1569.0 J / 10.0 °C = 156.9 J / °C

(50.0) (4.184) (7.5) = 1569.0 J.

1569.0 J / 10.0 °C = 156.9 J/°C

Determine a Calorimeter Constant I Determine a Calorimeter Constant II

q = m C p ΔT q = (72.55 g) (4.184 J g¯ 1 °C¯ 1 ) (24.3 °C) q = 7376.24 J

q = m C p ΔT q = (58.85 g) (4.184 J g¯ 1 °C¯ 1 ) (24.9 °C) q = 5818.54 J

7376.24 − 5818.54 = 1557.7 J

1557.7 J / 24.9 °C = 62.6 J/°C

q = (100.0 g) (18.8 °C) (4.184 J g¯ 1 °C¯ 1 ) = 7865.92 J

q = (100.0 g) (16.9 °C) (4.184 J/g °C) = 7070.96 J

7865.92 − 7070.96 = 794.96 J

794.96 J / 16.9 °C = 47.0 J/°C

## How to solve Calorimetry Problems

Coffee cup calorimeter - calculate enthalpy change, constant pressure calorimetry, sharing buttons:.

00:00 in this video we're gonna talk about how

00:03 to solve the coffee cup calorimeter

00:04 problem so let's consider this problem

00:07 that sauna board fifteen grams of

00:10 calcium chloride is dissolved in 250

00:14 milliliters of water in a coffee cup

00:16 calorimeter the temperature increases

00:18 from 25 point to Celsius to 35 point

00:22 seven and our goal is to calculate the

00:25 enthalpy change in kilojoules per mole

00:28 how can we do it well first let's draw a

00:31 picture of the coffee cup calorimeter so

00:36 basically what we have is a Styrofoam

00:38 cup styrofoam is a good thermal

00:41 insulator it's going to prevent heat

00:43 from flowing into or out of the coffee

00:47 and so what we're gonna have is some

00:49 water and we're gonna dissolve calcium

00:52 chloride into the solution and we're

00:59 gonna also have a thermometer that's

01:02 placed inside the Styrofoam cup now the

01:07 Styrofoam cup is sealed now you may need

01:15 to styrofoam cups to increase the

01:17 insulation so that no heat can go into

01:20 or out of it now ideally the Styrofoam

01:25 cup is not perfect but it works pretty

01:28 well as a good thermal insulator now the

01:33 reaction will either generate heat or

01:37 absorb heat and the temperature is going

01:40 to record the changes based on the

01:43 temperature of the surrounding solution

01:45 now here's the question for you if the

01:49 temperature of the solution increases do

01:52 we have an exothermic reaction or is the

02:01 now all of the heat generated by calcium

02:04 chloride will be absorbed by water or

02:08 the reaction can also absorb heat from

02:11 water which means water could really see

02:13 to it either case the heat absorbed or

02:17 released by the reaction is equal to the

02:20 heat absorbed or released by the

02:21 surrounding water molecules now we need

02:24 to put a negative sign on one side of

02:26 the equation in this problem the system

02:30 consists of the reactants and the

02:32 products the surroundings is basically

02:36 the water molecules so if the

02:39 temperature goes up that means water

02:43 absorb heat because the thermometer it

02:47 records the temperature of the water or

02:49 the solution so if the temperature goes

02:52 up the water molecules have more thermal

02:54 energy which means that they absorb team

02:58 so it's endo for the surroundings but

03:01 the system the reaction release heat so

03:05 that the water can absorb it so

03:08 therefore for the reaction it's

03:09 exothermic so if the temperature goes up

03:12 you have an exothermic reaction likewise

03:16 the reverse is true if the temperature

03:19 of the solution decreases the reaction

03:22 is endothermic the temperature tells you

03:28 what happens to the kinetic energy of

03:31 the water molecules so if the

03:32 temperature goes down that means that

03:35 the water molecules are losing thermal

03:37 energy which means they released it to

03:40 the reaction so it's exothermic for the

03:43 water but it's endothermic for the

03:46 reaction because the reaction it pulled

03:49 in heat energy from the surrounding

03:51 water molecules and so that's why the

03:54 temperature of the water molecules goes

03:55 down but the reaction itself the system

03:58 it's undergoing an endothermic process

04:03 so now looking at our example the

04:07 temperature of the solution goes up from

04:09 25 point two to thirty five point seven

04:13 which means that because the temperature

04:16 is going up the dissolution of calcium

04:18 chloride is an exothermic reaction so

04:22 Delta H of the reaction we should expect

04:25 to be negative now how do we go about

04:30 calculating the enthalpy change for this

04:33 reaction now if you notice the units we

04:39 need to be in kilojoules per mole

04:42 kilojoules is unit of energy so we got

04:45 to calculate Q of the reaction and we're

04:48 going to divide it by the number of

04:49 moles of this substance now we have the

04:53 mass of calcium chloride so we can use

04:55 the molar mass to convert it to moles so

04:59 how do we calculate Q of the reaction Q

05:02 of the reaction is equal to negative Q

05:06 of water so we need to calculate the

05:09 heat energy that's absorbed by all of

05:11 the water molecules and we can do so

05:13 using this equation it's equal to M C

05:17 delta T where m is the mass of the water

05:19 sample C is the specific heat capacity

05:22 of water which in this case is 4.184

05:26 joules per gram per Celsius and LC is

05:31 the change in temperature is the final

05:33 minus the initial temperature so we have

05:37 everything that we need in order to find

05:40 out 2h so let's go ahead and do it

05:46 so let's start with this equation let's

05:50 calculate the amount of thermal energy

05:51 absorbed by the 250 milliliters of water

05:55 now we need the mass in grams the

05:58 density of water is 1 gram per

06:00 milliliter so 250 milliliters of water

06:04 is equal to a mass of 250 grams in the

06:08 case of water the number of milliliters

06:10 is equal to the number of grams at this

06:14 density so 100 milliliters of water is

06:17 approximately 100 grams of water now the

06:20 density might slightly change based on

06:22 temperature but for all practical

06:24 purposes the density of water is about 1

06:30 now we have the mass which is 250 grams

06:33 we have the specific heat capacity which

06:36 is 4.184 joules per gram per Celsius and

06:41 the temperature change final minus

06:45 initial so if we take thirty five point

06:47 seven and subtracted by twenty five

06:50 points you that's going to give us a

06:52 temperature change of ten point five

06:55 degrees Celsius so notice that the unit

07:03 grams cancel and the unit Celsius

07:07 cancels as well leaving behind the unit

07:10 joules so let's multiply these three

07:13 numbers so it's 250 times four point one

07:17 eight four times ten point five so Q is

07:24 positive ten thousand nine hundred and

07:30 joules now keep in mind we need to get

07:36 the enthalpy change in kilojoules per

07:38 mole so that means we need to convert

07:40 Joules to kilojoules one kilojoule is

07:44 equal to a thousand joules so we got to

07:47 divide by a thousand in order to cancel

07:52 the unit joules so this is going to give

07:55 us ten point nine eight three kilojoules

08:00 now cue the reaction is negative q of

08:05 the water so right now we know that the

08:08 process is endothermic for the

08:11 surrounding water molecules which means

08:13 it must be exothermic for the system so

08:16 Q of the reaction is going to be

08:17 negative ten point nine eight three

08:20 kilojoules now all I need to do is need

08:24 to divide the kilojoules by the number

08:26 of moles of substance that we have so

08:31 we're dissolving 15 grams of calcium

08:34 chloride in order to convert this to

08:37 moles we need the molar mass so calcium

08:41 has an atomic mass of forty point zero

08:44 eight and chlorine is 35 point four five

08:48 don't forget to multiply by two based on

08:58 so the molar mass of calcium chloride is

09:00 1/10 0.98 grams per mole so one mole has

09:07 a mass of 110 point nine eight grams so

09:12 let's divide those two numbers and you

09:16 should get point one three five one six

09:20 moles of calcium chloride

09:33 now the last thing that we need to do is

09:36 calculate Delta H which we set it's Q of

09:40 the reaction divided by the number of

09:44 moles so that's gonna be negative ten

09:48 point nine eight three kilojoules

09:52 divided by point one three five one six

09:55 moles so you should get an answer that's

10:06 negative eighty one point three

10:09 kilojoules per mole so that's the

10:13 enthalpy change of this reaction based

10:16 on the data that we were given

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So the amount of heat used by the calorimeter to heat from 25 to 35 is. Example #3: A calorimeter is to be calibrated: 72.55 g of water at 71.6 °C added to a calorimeter containing 58.85 g of water at 22.4 °C

The calorimeter constants are used in constant pressure calorimetry to calculate the amount of heat required to achieve a certain raise in the temperature of the calorimeter's contents

The heat capacity of the calorimeter system in ( glucose). If heat capacity of calorimeter and its contents is , calculate the heat transferred to calorimeter

Constant-volume calorimeters, such as bomb calorimeters, are used to measure the heat of combustion of a reaction. A bomb calorimeter is a type of constant-volume calorimeter used to measure a particular reaction's heat of combustion

Mass of nested cups (two cups inside one another): 3.04 g Mass of single cups: 1.51 g 75 mL of cold water goes into nested cup: 76.045 g and the temp. remained constant at 23.0 75 ml of warm water into single cup: 72.755 g and the temp