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CALORIMETRIC ANALYSIS

1. Definethe  following terms

(a) Calorimetry

Calorimetry is the field of science that deals with the measurement of the state of a body with respect to the thermal aspects in order to examine its physical and chemical changes. The changes could be physical such as melting, evaporation or could also be chemical such as burning, acid-base neutralization etc. Calorimetry is applied extensively in the fields of thermochemistry in calculating the enthalpy, stability, heat capacity etc.

• Calorie
• Joule
• Calorific value

3.State the principle of calorimetry

The principle of calorimetry (or principle of mixtures) states that for an insulated system, heat energy lost by the hot body is equal to the heat energy gained by the cold body.

                        m1c1(t1−t)=m2c2(t−t2​)

Note: Heat transfer occurs until both the bodies attain the same temperature(t).

1. Describe a calorimeter

A calorimeter is a device used for heat measurements necessary for calorimetry. It mainly consists of a metallic vessel made of materials which are good conductors of electricity such as copper and aluminium etc. There is also a facility for stirring the contents of the vessel. This metallic vessel with a stirrer is kept in an insulating jacket to prevent heat loss to the environment. There is just one opening through which a thermometer can be inserted to measure the change in thermal properties inside. Let us discuss how exactly heat measurements are made. In the previous article, we discussed the specific heat capacity of substances.

Such measurements can be made easily with this. Say in a calorimeter a fixed amount of fuel is burned. The vessel is filled with water, and the fuel is burned, leading to the heating of water. Heat loss by the fuel is equal to the heat gained by the water. This is why it is important to insulate the calorimeter from the environment; to improve the accuracy of the experiment. This change in heat can be measured through the thermometer. Through such a measurement we can find out both the heat capacity of water and also the energy stored inside a fuel.

1. Define   the term enthalpy

• Enthalpy is the heat energy that is being absorbed or evolved during the progression of a chemical reaction. The enthalpyis given the symbol H. H indicates the amount of energy. The change of enthalpy is given as ∆H, and the symbol ∆ indicates the change of enthalpy. The heat of formation and heat of reaction are two forms of enthalpies.
  1. Explain the following terms
  2. Heat capacity

Heat capacity is an intrinsic physical property of a substance that measures the amount of heat required to change that substance’s temperature by a given amount. In the International System of Units (SI), heat capacity is expressed in units of joules per kelvin (J∙K−1). Heat capacity is an extensive property, meaning that it is dependent upon the size/mass of the sample. For instance, a sample containing twice the amount of substance as another sample would require twice the amount of heat energy (Q) to achieve the same change in temperature (ΔT) as that required to change the temperature of the first sample.

1. Specific Heat CapacitiesThe specific heat capacity, often simply called specific heat, is the heat capacity per unit mass of a pure substance. This is designated c_P and c_V and its units are given in Jg∙K.
2. Molar  heat capacity

The molar heat capacity  is the heat capacity per mole of a pure substance. Molar heat capacity is often designated CP, to denote heat capacity under constant pressure conditions, as well as CV, to denote heat capacity under constant volume conditions. Units of molar heat capacity are JK∙mol.

1. Thermochemical Equation is a balanced stoichiometric chemical equation that includes the enthalpy change, ΔH. In variable form, a thermochemical equation would look like this:

                    A + B → C

                       ΔH = (±) #

Where {A, B, C} are the usual agents of a chemical equation with coefficients and “(±) #” is a positive or negative numerical value, usually with units of kJ.

6.State Hess’s law of constant heat summation

Hess’s Law of Constant Heat Summation (or just Hess’s Law) states that regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes. This law is a manifestation that enthalpy is a state function

1. Explain
   1. Heat of formation

The heat of formation is the change of enthalpy when one mole of a compound is formed from its constituent elements. This is called standard heat of formation when it is given for the formation of substances at their standard states. However, there is no defined standard temperature. Hence, the heat of formation is given for a constant pressure. The symbol for the standard heat of formation is ΔH_f°. It can be calculated using the following equation.

According to this equation, the standard heat of formation is equal to the difference between the sum of the standard enthalpies of formation of the product and the standard enthalpies of formation of reactants.

• Heat of Reaction

The heat of reaction is the net amount of energy that should be added or released during a chemical reaction. In other words, it is the change in enthalpy in a chemical reaction that occurs at a constant pressure. The heat is either added or released in order to keep all the reactants at a constant temperature. It is measured per mole of a substance. The heat of reaction is given as below. When this is defined for the standard state, it is known as standard enthalpy of reaction. The standard state is the pure substance at 1 bar pressure and at a relevant temperature.

If the value of the heat of reaction is a positive value, it indicates that the reaction is endothermic (in which the heat is absorbed from outside). But if the value of the heat of reaction is a negative value, it indicates that the reaction is exothermic (in which heat is given off).      

2. Distinguish between heat of formation and heat of reaction

The main difference between heat of formation and heat of reaction is that heat of formation is the amount of energy either absorbed or released during the formation of a compound whereas heat of reaction is the amount of energy either absorbed or released during any chemical reaction.

3. State and explain  Hess law

Hess’s law is useful for  determining the overall  change in enthalpy  especially in a reaction which involves two or more parts.. It states that the enthalpy change for a reaction or process is independent of the route through which it occurs. This means that if reaction transforms a substance into another, it doesn’t matter if the reaction occurs in one step (reactants become products immediately) or whether it goes through many steps (reactants become intermediaries and then become products), the resulting enthalpy change is the same in both cases.

Hess’ law states that the change in enthalpy of the reaction is the sum of the changes in enthalpy of both parts. In this case, the combustion of one mole of carbon has ∆H = −394 kJ/mol (this happens six times in the reaction), the change in enthalpy for the combustion of one mole of hydrogen gas is ∆H = −286 kJ/mol (this happens three times) and the carbon dioxide and water intermediaries become benzene with an enthalpy change of ∆H = +3,267 kJ/mol.

Take the sum of these changes to find the total enthalpy change, remembering to multiply each by the number of moles needed in the first stage of the reaction:

                  ∆H_total = 6×(−394) + 3×(−286) +3,267

                             = 3,267 − 2,364 – 858

                            = 45 kJ/mol

The enthalpy change is negative because the reaction releases heat to the surroundings, resulting in an increase in temperature of the water.

4. A 50g sample of a metal was heated to 95oCthen quickly transferred to an insulated container containing 50g of H2O at 25oC.  The final temperature of the H2O was 30oC.   What can be concluded ?

The specific heat of the water is greater than that of the metal. This is because when the heated metal is placed in the container of the cooler water there will be a transfer of thermal energy from the metal to the water.  This transfer will occur towards an equilibrium of thermal energy in the water and in the metal. Thus we can conclude that the amount of thermal energy lost by the metal will equal the amount of thermal energy gained by the water.  However we notice that the water increases by only 5^oC  and the metal decreases by 65^oC.  This is becasue of the difference of the specific heats of these substances.  The specific heat capacity of a substance is the heat required to increase the temperature of 1g of a substance by 1^oC. The metal can be conluded to have a smaller specific heat than the water because the same amount of energy transfer led to a much larger change in termperature for the metal as compared to the water. 

5. How much energy is needed to raise the temperature of five grams of ice from −10^oCto 35^oC?   This question involves the total energy needed for three different processes: the temperature raise from −10oC to 0oC, the melting of the ice, and the temperature raise from 0oC to 35oC. For the first and third transitions we will use the equation q=mCΔT. For the melting of ice, we will use the equation q=mΔHfusion.
6. q=miceCiceΔT=(5g)(0.5cal/g∘C)(10∘C)=25cal
7. q=mΔHfusion=(5g)(80cal/g)=400cal
8. q=mwaterCwaterΔT=(5g)(1.0cal/g∘C)(35∘C)=175cal

Finally, we will need to sum the energy required for each step to find the total energy.

                  25cal+400cal+175cal=600cal

6. A 20g sample of iron at a temperature of 120^oCis placed into a container of water. There are 300 milliliters of water in the container at a temperature of 30^oC. 

Ciron=0.444Jg^∘C    Cwater=4.184Jg^∘C   ρwater=1gmL

How much heat does it take to heat 100g ice at 0C to boiling point?

You need heat for the phase change, using the enthalpy of fusion (100g X 334 J/g = 33400 J). Add to this the heat to get to boiling point using the specific heat of water

               =     100g X 100C X 4.2 J      

                        = 42000 J).

                               =  33400 J +   42000J   

                                          = 75400 J

                             Totalling =     (75.4 kJ)

What is the final temperature of the water?

There are two things to note before solving for the final temperature.

1. The density of water allows us to say that 300 milliliters of water is the same thing as 300 grams of water.

300mlx 1g1mL=300g

2. Since the heat from the iron is being transferred to the water, we can say that the heat transfer is equal between both compounds. Since the heat is conserved in the system, we can set the two equations equal to one another.

q=mCΔT

qi=qw→miCiΔT=mwCwΔT

miCi(T_initial−T_final)=mwCw(T_final−T_initial)

Notice how the change in temperature for iron has been flipped in order to avoid a negative number.

(20g)(0.444J/g^∘C)(120^∘C−T_f)=(300g)(4.184J/g^∘C)(T_f−30^∘C)

T_f=30.63∘C

Because water has a much higher heat capacity compared to iron, the temperature of the water is not changed significantly.

7. Calculate ΔH for the following reaction:

CH_{4 (g)} + O₂ (g)  ⇌  CO_{2 (g)} + H₂O (l)

Compound                ΔH

CH₄ (g)             -74.8 kJ/mol

H₂O (l)             -285.8 kJ/mol

CO₂ (g)            -393.5 kJ/mol

ΔH = Σ ΔHf products – Σ ΔHf reactants

ΔHf O₂ or any element is 0

First step is to balance the equation:

CH_{4 (g)} + O₂ (g)  ⇌  CO_{2 (g)} + 2H₂O (l)

ΔH = Σ ΔHf products – Σ ΔHf reactants

= [-393.5 kJ/mol + 2(-285.8) kJ/mol] – (-74.8 kJ/mol)

= -890.3 kJ/mol

8. What mass of carbon monoxide must be burned to produce 175 kJ of heat under standard state conditions?

Given that:

          CH4 (g)      C (s) + 2 H2 (g) ΔHr = 74.8 kJ mol¹

What is the ΔHf of CH4 (g)?

                 –74.8 kJ mol^–1

The required reaction is the reverse of formation, so as we are going in the opposite direction, it requires the opposite sign.

9. A metal weighing 4.82 g was heated to 115.0 °C and put into 35 mL of water of temperature 28.7 °C. The metal and water were allowed to come to an equilibrium temperature, determined to be 34.5 °C. Assuming no heat was lost to the environment, calculate the specific heat of the metal. Consider the specific heat capacity of water as 4.186 joule/gram °C.

First, let us calculate the heat absorbed by the water and use the value obtained to calculate the specific heat of the metal.

The formula to find the heat absorbed by the water is given as

           q=mcΔt

Substituting the values in the equation, we get

qabsorbed=(4.186J/g⋅∘C) × 35g × (5.8  ^oC)

                  = 850J

Now, using this formula let us calculate the specific heat of the metal as follows:

850J=s×4.82g×80.5 ^oC

s=2.19J/G⋅ ^oC

The specific heat of the metal is 2.19J/G⋅∘C

The molar heat capacity of water, CP, is 75.2 Jmol∙K.

10. How much heat is required to raise the temperature of 36 grams of water from 300 to 310 K?

We are given the molar heat capacity of water, so we need to convert the given mass of water to moles:

 

36 grams×1 mol H2O 

                             18 g                =   2.0 mol H2O

Now we can plug our values into the formula that relates heat and heat capacity:

q=nCPΔT

q=(2.0mol)(75.2Jmol∙K)(10K)

q = 1504 J

11. A solution was made by dissolving a spatula of potassium nitrate into 50 cm3 of water. The temperature changed from 20.4˚C to 18.7˚C. Calculate the enthalpy change for this reaction.

To calculate the enthalpy change (ΔH) we must know the values for c, m and ΔT. The specific heat capacity (c) is a constant, with a value of 4.18. Since 50 cm³ of water have been used, the mass of water (m) is 0.05 kg.

From the question we can see that the temperature has decreased by 1.7 ˚C. This means that the reaction is endothermic (so ΔH will be positive).

ΔH=cmΔT

= 4.18 x 0.05 x 1.7     = 0.3553 kJ

In an experiment on the specific heat of a metal a 0.20 kg block of the metal at 150oC is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150cm3 of water at 27oC. The final temperature is 40oC. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value for specific heat of the metal ? 

Mass of the metal, m = 0.20 kg = 200 g 

Initial temperature of the metal, T1​ = 150 ^oC

Final temperature of the metal, T2​ = 40 ^oC

Calorimeter has water equivalent of mass, m = 0.025 kg = 25 g

Volume of water, V = 150 cm3

Mass (M) of water at temperature T = 27oC:

150×1=150g

Fall in the temperature of the metal:

ΔTm​=T1​-T2​ =150−40=110^oC
Specific heat of water, Cw​=4.186J/g/K

Specific heat of the metal =C

Heat lost by the metal,  =mCT …. (i)

Rise in the temperature of the water and calorimeter system:                                        

                  T1​−T=40−27=13^oC
Heat gained by the water and calorimeter system: 

      =m1​Cw​T=(M+m)CwT ….(ii)

Heat lost by the metal = Heat gained by the water and colorimeter system

                mCΔTm​=(M+m)Cw​Tw​

200×C×110=(150+25)×4.186×13

C=(175×4.186×13)/(110×200)

=0.43Jg⁻1k⁻¹

If some heat is lost to the surroundings, then the value of C will be smaller than the actual value.

12. A 360-g piece of rebar (a steel rod used for reinforcing concrete) is dropped into 425 mL of water at 24.0 °C. The final temperature of the water was measured as 42.7 °C. Calculate the initial temperature of the piece of rebar. Assume the specific heat of steel is approximately the same as that for iron  and that all heat transfer occurs between the rebar and the water (there is no heat exchange with the surroundings).

Solution
The temperature of the water increases from 24.0 °C to 42.7 °C, so the water absorbs heat. That heat came from the piece of rebar, which initially was at a higher temperature. Assuming that all heat transfer was between the rebar and the water, with no heat “lost” to the surroundings, then  heat given off by rebar = −heat taken in by water, or:

      qrebar=−qwater

Since we know how heat is related to other measurable quantities, we have:

         (c×m×ΔT)rebar=−(c×m×ΔT)water

Letting f = final and i = initial, in expanded form, this becomes:

Crebar×mrebar×(Tf,rebar−Ti,rebar)=−cwater×mwater×(Tf,water−Ti, water)

The density of water is 1.0 g/mL, so 425 mL of water = 425 g. Noting that the final temperature of both the rebar and water is 42.7 °C, substituting known values yields:

(0.449J/g^∘C)(360g)(42.7^∘C−Ti,rebar)=(4.184J/g^∘C)(425g)(42.7^∘C−24.0^∘C)

 Ti, rebar= (4.184J/g^∘C)(425g(42.7^∘C−24.0^∘C)=42.7^∘C (0.449J/g∘C)(360g)  

Solving this gives T_i,rebar= 248 °C, so the initial temperature of the rebar was 248 °C.

13. Calculate the ΔHf of methane (CH4 (g)), using the following ΔHC data: CH4 (g) = –882 kJ mol^–1; C (s) = –394 kJ mol^–1; H2 (g) = –286 kJ mol^–1

Step I; Construct a Hess’s Law cycle:

C (s) + 2 H2 (g) CH4 (g)+ 2 O2 (g)                    2 O2 (g)

 

                                                   CO2 (g) + 2 H2O (l)

We want ΔHf which is the reaction going across the top (blue arrow). To get there we go down the black arrow and back up the orange arrow. We travel in the direction of the black arrow, but in the opposite direction for the orange arrow.

So, we get:

    C (s) + 2 H2  (g) + 2 O2 (g)                             CO2 (g) + 2 H2O (l)

       – CH4  (g) + 2 O2 (g)                               CO2 (g) + 2 H2O (l)

= (–394 + 2 x (–286)) – (–882) kJ mol^–1

= –84 kJ mol^–1

41.Explain why instant cold packs used by athletes in managing muscular injuries  contain ammonium nitrate and water.

Cold pack is  a Popular and effective in treatments to ease pain and swelling from minor injuries, cold packs come in many different varieties. Some are sacks of gel that turn into ice packs in your freezer; others are packets designed to turn cold instantly with a simple squeeze, no refrigeration or freezing required. You can also make your own cold pack by wrapping some

Anyone who has ever sprained or twisted an ankle or pulled a muscle knows that cold is your friend. Bruises, insect bites, and repetitive strain injuries such as tendinitis, also respond well to treatment with cold packs. Cold therapy can help people with muscle spasms, whiplash, and various forms of arthritis as well.

Cold packs are very effective at reducing swelling and numbing pain. An injury swells because fluid leaks from blood vessels; cold causes vessels to constrict, reducing their tendency to ooze. The less fluid that leaks from blood vessels, the less swelling results. Cold also eases inflammation and muscle spasms, two common sources of pain. cold pack has a plastic bag containing water. Inside this bag is a smaller bag

containing ammonium nitrate. The outer bag is squeezed so that the inner bag bursts. The pack is shaken and quickly gets

very cold as the ammonium nitrate dissolves in the water according to the endothermic reaction.  NH₄NO₃(s) → NH₄⁺ (aq) + NO⁻³⁽aq)

The ammonium nitrate produces lower mean minimum temperatures, and reaches them more quickly, demonstrating it is a more effective chemical for use in relieving pain.

The sooner you apply an ice pack to a sprain or strain, the sooner it can do its job reducing pain and swelling. For chronic problems such as low back pain or muscle spasms, ice whenever the symptoms start up.

A general rule of thumb is to ice an injury over a period of 24 to 72 hours. Apply cold packs for periods of up to 20 minutes every two to four hours. When your skin starts to feel numb, it’s time to give your body a break from a cold pack.However ,prolonged, direct contact with cold can damage skin and nerves so always be sure to wrap your cold pack in a towel. Distinguish between exothermic and endothermic reaction