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 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.
- "Chemical Principles: The Quest for Insight"; Peter Atkins, et al.; 2008
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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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