Calorimetry

Calorimetry is a method by which heat transfer into and out the system during the chemical reactions (combustion reactions for example) is measured. In this process the heat transfer is calibrated with the calibrated object. The temperature change in the process is used to calculate the heat transferred. This calculation needs to be specific based on surrounding and system definitions.

System: the chemical compound which is undergoing chemical reaction.

Surrounding: the calorimeter with ALL its components including the equipment that measures the heat.

Calorimeter:

Calorimeter is a tool or equipment used to measure the heat absorbed (endothermic: temperature will decrease) or given away by (exothermic: temperature will increase) during a chemical reaction.

There are five types of calorimeters in use:

There are 2 major types of calorimeters:

1. Adiabatic Calorimeters
2. Reactions Calorimeters

1. Adiabatic Calorimeters:

These types of Calorimeters are used to measure the enthalpy change in a system, during physical processes such as crystallization, mixing, dilution etc. These calorimeters do not exchange heat with the surround and function at zero heat exchange conditions also known as adiabatic conditions.

The video below illustrates the function of the adiabatic calorimeter:

Adiabatic calorimetry

Figure 5.13 Adiabatic Calorimeter

2. Reactions Calorimeters

There are several reactions calorimeter available in the market and in the universities and colleges

These calorimeters measure the heat exchange or transfer in exothermic and endothermic reactions. The reactions calorimeters can measure chemical or physical reactions or changes.

Types of Reactions Calorimeters:

4 types of reactions calorimeters are available

Heat-flow,

Heat-balance,

Constant flux

Power compensation

Detailed functionalities regarding these types can be read at:

                  Heat Flow Calorimeters examples are given below:

1. Foam Cup Calorimeter:

This type of calorimeter is used mostly at undergraduate physics and chemistry labs because it cheap and easy to use and has reasonable % error (but not very accurate). The two figures below show the setup of this foam cup calorimeter.

Inside the thermometer is specific amount of water with specific volume (using the density of water this volume is converted into mass of water). A weighed metal of liquid solution is warmed up to 50^oC to 60^oC and then immediately transferred into the foam cup (calorimeter) and the temperature is followed till there is no rise in the temperature and an equilibrium temperature of the system is attained.

Foam cup calorimeter

Figure 5.14 Styrofoam Cup Calometer

Figure 5.15 Coffee Cup Calorimeter Diagram

Reference: https://commons.wikimedia.org/wiki/File:Coffee_cup_calorimeter_pic.jpg

Figure 2 Reference:http://wiki.chemprime.chemeddl.org/articles/a/_/c/File~A_Coffee_Cup_Calorimeter_.jpg_6993.html

2. Bomb  Calorimeter:

This type of calorimeter is used in commercial applications as well as in research labs at the universities and special institutes. Bomb calorimeter is used to measure the heat of combustion reaction. It uses stainless steel combustion reaction chamber with the presence of oxygen. Immediately after the combustion reaction occurs which is assisted with an electrical current goes through the samples to ignite for few seconds, the heat is transferred to the water which is contained in outer sealed mantel of the combustion chapter. The heat will warm up the water and the temperature is monitored and recorded and the heat of the combustion is calculated.

Bomb Calorimeter

Figure 5.16 Bomb Calorimeter

Reference: https://commons.wikimedia.org/wiki/File:Bomb_Calorimeter_Diagram.png

The videos below illustrates the bomb calorimeter concept:

Bomb Calorimeter vs Coffee Cup Calorimeter Problem – Constant Pressure vs Constant Volume Calorimet

Bomb Calorimeter

3. Differential Scanning Calorimeter:

This type of calorimeter is used to test the physical properties of plastics, polymers, monomers and elastomers. This calorimeter deals with the thermal analysis technique in the research and quality control.  The calorimeter monitors the product analysis at different range of temperatures and heat changes which facilitates the study of the chemical properties of the sample within its heat capacity.

The videos below illustrate the concept of the Differential Scanning Calorimeter:

What Is Differential Scanning Calorimetry (DSC)?

Differential Scanning Calorimetry

4. Solution Calorimeter:

This type of calorimeter is used to measure the heat of solution of dissolving a solid compound in in a liquid solution. Also the heat of solution can be called the heat of dilution when dissolving a liquid solution in another liquid solution.

The videos below illustrate the concept of the Solution Calorimeter:

Solution Calorimetry

Solution Calorimetry Pchem1

The ice calorimeter is an important tool for measuring the heat capacities of liquids and solids, as well as the heats of certain reactions. This simple yet ingenious apparatus is essentially a device for measuring the change in volume due to melting of ice. To measure a heat capacity, a warm sample is placed in the inner compartment, which is surrounded by a mixture of ice and water.

The heat withdrawn from the sample as it cools causes some of the ice to melt. Since ice is less dense than water, the volume of water in the insulated chamber decreases. This causes an equivalent volume of mercury to be sucked into the inner reservoir from the outside container. The loss in weight of this container gives the decrease in volume of the water, and thus the mass of ice melted. This, combined with the heat of fusion of ice, gives the quantity of heat lost by the sample as it cools to 0°C.

Figure 5.19 Ice Calorimeter

The Heat of Solution, Heat of Dilution and Heat Neutralization as well as Specific Heat Calculations using Heat Flow Calorimeters:

The temperature is followed till there is no rise in the temperature and an equilibrium temperature of the system is attained.

Then general formula is: (at thermal equilibrium at the end of reaction or solution or dilution).

q(compound A) + q(compound B) = 0

q(compound A)  =  –  q(compound B)

-Heat is lost by compound A = Heat is gained by compound B

The heat flow is from A → B

Examples:

Heat transfer or exchange between two substances (similar to the foam cup calorimeter).

1. A piece of metal weighing 59.047 g was heated to 100.0 °C and then put it into 100.0 mL of water (initially at 23.7 °C). The metal and water were allowed to come to an equilibrium temperature, determined to be 27.8 °C. Assuming no heat lost to the environment, calculate the specific heat of the metal.

-Heat is lost by the metal = Heat is gained by water

-mass_(metal) x c_(metal) x ∆T_(metal)   = mass_(water) x c_(water) x ∆T_(water)

-[59.047 g] x c_(metal) x [100 ^oC – 27.8 ^oC]  = [100.0 mL x 1.00 g / mL] x [4.184 J / (g x ^oC)] x [27.8 ^oC – 23.7 ^oC]

[59.047 g] x c_(metal) x [100 ^oC – 27.8 ^oC] = 1713.8 J

c_(metal)  = [1713.8 J] / { [59.047 g] x [100 ^oC – 27.8 ^oC] } = 0.402 J/g x ^oC

2. In a coffee-cup calorimeter, 100.0 g of H₂O and 100.0 mL of HCl are mixed. The HCl had an initial temperature of 44.6 ^oC and the water was originally at 24.6 ^oC. After the reaction, the temperature of both substances is 31.3 ^oC.

1. Was the reaction exothermic or endothermic? Explain.

For the water- endothermic. The temperature increased from 24.6 ^oC to 31.3 ^oC indicating energy was absorbed by the water.

For the HCl- exothermic. The temperature decreased from 44.6 ^oC to 31.3 ^oC indicating energy was released by the HCl.

2. Calculate how much heat the water lost or gained.

q(_water) = mass_(water) x c_(water) x ∆T_(water) = [100.0 g] x [4.184 J / (g x ^oC)] x [31.3 ^oC – 24.6 ^oC] =

q(_water) = 2803.3 J  =  2.80 x 10³ J   = 2.80 kJ

3. What is the final temperature after 840.0 Joules is absorbed by 10.0g of water at 25.0^oC?

q(_water) = mass_(water) x c_(water) x ∆T_(water)

840.0 J = [10.0 g] x [4.184 J / (g x ^oC)] x ∆T_(water)

∆T_(water) =   [[840.0 J] / { [10.0 g] x [4.184 J / (g x ^oC)] = 20.1 ^oC

T_(Final) = 25.0 + 20.1 =  45.1 ^oC

4. A 0.88 g gummy bear is burned in a bomb calorimeter. The temperature started at 21.5 °C and leveled off at 24.2 °C. The manufacturer of the bomb calorimeter determined the heat capacity of the calorimeter to be 11.4 kJ/°C. Calculate the heat of combustion per gram of gummy bear.

q = c_{(calorimeter)} ΔT = [11.4 kJ/°C] x [24.2-21.5 °C] = [11.4 kJ/°C)(2.7 °C] = 30.78 kJ

(30.78 kJ) / (0.88 g) = 34.98 kJ / g    

5. A piece of metal weighing 85.5 g was heated to 100.0 °C and then put it into 100.0 mL of water (initially at 23.7 °C). The metal and water were allowed to come to an equilibrium temperature, calculate this final temperature (the equilibrium temperature) if the specific heat of the piece of the metal is 0.350 J / (g x °C)

-Heat is lost by the metal =  Heat is gained by water

-mass_(metal) x c_(metal) x ∆T_(metal)   = mass_(water) x c_(water) x ∆T_(water)

[85.5 g] x [0.350 J / (g x °C)] x [T_(final) – T_{(initial metal)}]  = – [100.0 mL x 1.00 (g / mL)] x [4.184 J  / (g x ^oC)] x [T_(final) – T_{(initial water)}]

[29.925 / °C] x [T_(final) – 100.0 °C] = – [418.4  / °C] x [T_(final) – 23.7 °C]

29.925 T_(final)  – 2992.5  =  – 418.4 T_(final)  +  23.7]

29.925 T_(final)  +   418.4 T_(final)    =    23.7 + 2992.5 = 3016.2

448.325 T_(final) =  3016.2

T_(final) = [3016.2 / 448.325] = 6.7277 °C = 6.73 °C

6. A 0.258 g piece of potassium nitrate solid is placed into water inside of a coffee cup calorimeter resulting in a vigorous reaction. Assume a total volume of 100 mL for the resulting solution. The temperature of the solution changes from 22 °C to 25.1 °C due to the reaction. How much heat is generated per gram of potassium for this reaction? Assume the density of the solution after the reaction is the density of water and that the heat capacity of the solution and reaction vessel is only due to the water that has a specific heat of 4.184 J/ (g x °C).

-Heat is lost by the metal = Heat is gained by water

q = m x c_(solution) x ΔT_(solution) = [100 g] x [(4.184 J/g-°C)]  x [25.1 °C  – 22 °C]  = 1297 J

q = m x c_(solution) x ΔT_(solution) = [100 g] x [(4.184 J/g-°C)]  x  [3.1 °C]

q = (1297 J)/(0.258 g) = 5027 J/ g

7. 25.0 mL of 0.15 M HCl is neutralized with 30.0 mL of 0.25 M inside a foam cup calorimeter. Both acid and base were initially at 23.5 °C when they were added to the foam cup calorimeter. The final temperature recorded for this neutralization is 26.5 °C. Calculate the heat of neutralization reaction.

HCl(aq) + NaOH(aq)  ⟶  NaCl(aq) +  H₂O(l)

The heat taken by the solution = -heat given off by reaction

-q_(solution)  =  q_(reaction) = q_{(neutralization)}

q_(solution)  =  m x c_(solution) x ΔT

1. Assume that the specific heat of reaction equals the specific heat of water (water is the solvent for both acid and base).
2. Furthermore, the density of the solution is assumed to be the density of water 1.00 g / mL

The total volume of the solution = Volume of HCl + Volume of NaOH = 25.0 mL + 30.0 mL = 55.0 mL

3. Mass of solution =  [55.0 mL x  1.00 g / mL] = 55.0 g
4. ΔT = [26.5  °C – 23.5  °C]  = 3.0  °C

q_(solution)  =  q_{(neutralization)} =  [55.0 g] x [ [4.184 J/ (g x °C)] x [3.0 °C] =690.36 J = 690 J

8. A 1.5 kg block of Ni at 100 °C is placed into 500 mL of water that has a temperature of 21 °C. What is the final temperature assuming the specific heat of Ni is 0.44 J/g-°C and the specific heat of water is 4.184 J/g-°C. Hint: the total heat lost is equal to the total heat gained

Heat is lost by the Ni metal = – Heat is gained by water

q_(metal)   =  –  q_(water)

[1500 g Ni] x [0.44 J / g x °C] x [100 °C – T_f]  =  [500 mL x  1g/1mL] x [4.184 J / g x °C] x [T_f – 21 °C]

Re-arranging:

T_f =  40 °C

9. 25.6 g of ethanol was combusted in a bomb calorimeter and the heat obtained is that has a heat of 6.53 kJ . The initial temperature of the combustion was at 24.0 °C. Find its final temperature.

C₂H₅OH   +    3 O₂      ….>              2 CO₂     +    3   H₂O

-Heat is lost by the Ni metal = Heat is gained by water

-q_(ethanol)   =   q_(water) = q_{(heat of reaction)} = q_(combustion) = m x  c_(ethanol)  x   ΔT

Assuming the specific heat is equal to heat of water = 4.184 J / g x °C

6.53 kJ = 6.53 x 1000 J =  – [25.6 g] x [4.184 J / g x °C] x [T_(final)  –  T_(initial)]

6.53 kJ   =  6.53 x 1000 J  = – [107 J  /  °C] x [[T_(final)  –  24.0 °C]

6.53 x 1000 J   = [(107 J  /  °C)] T_(final) + 2568 J = (107 J  /  °C) T_(final)  + 2568 J

6.53 x 1000 J   –  2568 J  =  [(107 J  /  °C)] T_(final

3962 J  =   (107 J /  /  °C) T_(final)

T_(final₎  =  37.0 °C

10. When a 100. g sample of methane, CH₄, is burned in a bomb calorimeter the temperature changes from 21.0 °C to 31.0 °C and 2200. J of heat is given off. What is the specific heat of methane?

q_(combustion)  = m x c x ΔT  

c = q_(combustion)  /  m x ΔT = [2200 J] x /  [100. g  x 10.0 °C] 

Specific heat =  q_(combustion)  /  m x ΔT    =  [2200. J   /  °C] x  /  [100.  g] = 2.20 J / g x °C

11. When 3.12 g of glucose, C₆H₁₂O₆, is burned in a bomb calorimeter, the temperature of the calorimeter increases from 23.8 °C to 35.6 °C. The calorimeter contains 775 g of water, and the bomb itself has a heat capacity of 893 J/°C. How much heat was produced by the combustion of the glucose sample?

The heat produced by the reaction is absorbed by the water and the bomb:

qrxn = −(qwater + qbomb)

qwater  = [775 g] x [(4.184 J / g x ^oC)] x [35.6 ^oC – 23.8 ^oC] = 38262.68 J

qbomb  = [893 J / ^oC] x [35.6 ^oC – 23.8 ^oC] = 10537.4 J

Add both (qwater + qbomb) = 48.8 J

This reaction released 48.8 kJ of heat when 3.12 g of glucose was burned.

The following activity has been taken from Chemcollective.org

http://chemcollective.org/activities/autograded/115

Activity:

Coffee

During the summer after your first year at Carnegie Mellon, you are lucky enough to get a job making coffee at Starbucks, but you tell your parents and friends that you have secured a lucrative position as a “java engineer.” An eccentric chemistry professor (not mentioning any names) stops in every day and orders 250ml of Sumatran coffee at precisely 75.0°C. You then need to add enough milk at 5.00°C to drop the temperature of the coffee, initially at 80.0°C, to the ordered temperature.

Calculate the amount of milk (in ml) you must add to reach this temperature. Show all your work in the provided spaces.

In order to simplify the calculations, you will start by assuming that milk and coffee have the specific heat and density as if water. In the following parts, you will remove these simplifications. Solve now this problem assuming the density is 1.000 g/ml for milk and coffee and their specific heat capacity is 4.184 J/(g ºC).

Hint: the coffee is in an insulated travel mug, so no heat escapes. To insulate a piece of glassware in Virtual Lab, Mac-users should command-click (or open-apple click) on the beaker or flask; Windows users should right click on the beaker or flask. From the menu that appears choose “Thermal Properties.” Check the box labeled “insulated from surroundings.” The temperature of the solution in that beaker or flask will remain constant.

Use the Virtual Lab to verify your answer.