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

00:46 cup calorimeter

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

01:56 reaction endothermic

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

07:29 eighty-three

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

08:51 the subscript

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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How to Solve for Final Temperature in a Calorimeter

how to solve calorimeter constant problems

How to Find Heat Reaction When Zn Reacts With HCl

With a calorimeter, you can measure reaction enthalpies or heat capacities using the final temperature (Tf) of the contents. But what if you know the reaction enthalpy of your reaction and the heat capacities of the materials you are using and you want to predict what Tf will be instead? You can do this too — and in fact, this kind of problem is a common question on quizzes in chemistry classes.

Reread the homework/quiz question and determine what information you can extract from the question. You will probably be given a reaction enthalpy, the calorimeter constant and the heat capacity of the mixture formed by the reaction in the calorimeter, together with the starting temperatures.

Assume the calorimeter is perfect, i.e. that it does not lose heat to its environment.

Remember that in a perfect calorimeter, the heat given off by the reaction is equal to the sum of the heat gained by the calorimeter and the heat gained by its contents. Moreover, both the calorimeter and its contents will reach the same final temperature -- Tf. Consequently, you can use this information to write the following equation: Reaction enthalpy = (heat capacity of contents) x (mass of contents) x (Ti - Tf) + (Calorimeter constant) x (Ti - Tf) where Ti is the initial temperature and Tf is the final temperature. Notice that you're subtracting Tfinal from Tinitial and not the other way around. That's because in chemistry, reaction enthalpies are negative if the reaction gives off heat. If you want, you can subtract Ti from Tf instead, as long as you remember to flip the sign on your answer when you're done.

Solve for Tf as follows: Reaction enthalpy = (heat capacity of contents) x (mass of contents) x (Ti - Tf) + (Calorimeter constant) x (Ti - Tf)

Factor (Ti - Tf) out of the right side to yield: Reaction enthalpy = (Ti - Tf) x ( (heat capacity of contents) x (mass of contents) + (Calorimeter constant) )

Divide both sides by ( (heat capacity of contents) x (mass of contents) + (Calorimeter constant) ) to yield the following: Reaction enthalpy / ( (heat capacity of contents) x (mass of contents) + (Calorimeter constant) ) = Ti - Tf

Flip the sign on both sides then add Ti to both sides to yield the following: Ti - ( Reaction enthalpy / ( (heat capacity of contents) x (mass of contents) + (Calorimeter constant) ) ) = Tf

Plug in the numbers given you as part of the question and use them to calculate Tf. For example, if the reaction enthalpy is -200 kJ, the heat capacity of the mixture formed by the reaction is 0.00418 kJ/gram Kelvin, the total mass of the products of the reaction is 200 grams, the calorimeter constant is 2 kJ / K, and the initial temperature is 25 C, what is Tf?

Answer: First, write out your equation: Tf = Ti - ( Reaction enthalpy / ( (heat capacity of contents) x (mass of contents) + (Calorimeter constant) ) )

Now, plug in all your numbers and solve: Tf = 25 degrees - (-200 kJ / (0.00418 kJ/g K times 200 g + 2 kJ/K) ) Tf = 25 degrees - (-200 kJ / 2.836 kJ/K) Tf = 25 + 70.5 Tf = 95.5 degrees C

Things You'll Need

Related articles, how to calculate heat absorption, how to calculate the amount of heat released, how to calculate the mass of reaction in a mixture, how to convert kilojoules to kilocalories, how to calculate joules of heat, how to calculate energy released & absorbed, how to calculate heat absorbed by the solution, what is the unit for enthalpy, how to calculate a final temperature, how to calculate the molar heat of neutralization, how to mix calcium chloride and water, how to calculate kf, how to dissolve magnesium chloride, how is the equilibrium constant of a reaction determined, how to calculate btu for heat, how to calculate heat of sublimation, how to calculate hco3 from co2, how to determine a calorimeter constant, how to calculate molar heat capacity.

About the Author

Based in San Diego, John Brennan has been writing about science and the environment since 2006. His articles have appeared in "Plenty," "San Diego Reader," "Santa Barbara Independent" and "East Bay Monthly." Brennan holds a Bachelor of Science in biology from the University of California, San Diego.

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I need to find the calorimeter constant for this experiment.

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. dropped from 64.75 to 61.25 75 mL of warm water and 75 mL of cold water are placed into the nested cups and a cup is placed overtop of the nested cups to create the calorimeter. The change in temp. for the hot and cold water mixed was 38.5 to 40.75

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  5. I need to find the calorimeter constant for this experiment. Mass of nested cups (two cups

    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