Heat Capacity and Specific Heat Capacity
It can be observed from everyday experience that when a substance is warmed the temperature of the substance will
increase and when a substance is cooled the temperature of the substance will decrease. When a substance is
warmed, heat is absorbed, that is to say heat is transferred into the substance. When a substance is cooled, heat is
released, that is to say heat is transferred out of the substance.

Heat capacity (C) is the amount of heat (q) required to raise the temperature of an object one degree Celsius.  The units for heat capacity are J/^oC (the unit is read as Joules per degree Celsius). The equation which describes this relationship is: C = q/DT

The change in temperature (DT) of the object is defined as: DT = T_final - T_initial
 

Example If 95.0 J of heat is supplied to a sample of brass and the temperature of the sample increases by 17.1 oC, what is the heat capacity of this sample of brass? C = q/DT = 95.0 J/17.1 oC = 5.56 J/oC

Specific heat capacity (specific heat) is the amount of heat required to raise the temperature of one gram of a substance by one degree Celsius.  The value of the specific heat of water is 4.184 J/g×^oC  (the unit is read as Joules per gram per degree Celsius).

If the mass and specific heat of a substance is known, then the change in the substance’s temperature (DT) will indicate the amount of heat (q) that has been absorbed or released in a particular process. The equation for calculating the heat transferred is:

    q = (mass of the substance)(specific heat of the substance)(DT)

The sign associated with q will depend on DT. When the final temperature is higher than the initial temperature, the value of DT will have a positive sign (+). Thus, the sign of the value of q is positive (+q). This is consistent with the sign convention for an endothermic process wherein heat flows into a substance. When the final temperature is lower than the initial temperature, the value of DT will have a negative sign (-). Thus, the sign of the value of q is negative (-q). This is consistent with the sign convention for an exothermic process wherein heat flows out of a substance.
 

Example A 25.0 g sample of water is heated from 27.0oC to 53.0oC. Calculate the amount of heat absorbed by the water. q = (mass)(specific heat)(DT) = (25.0 g)( 4.184 J/g×oC)( 53.0oC - 27.0oC) = 2720 J = 2.720 kJ

Note: Heat capacity = (mass of substance)(specific heat of substance)

Calorimeter
A calorimeter is an apparatus designed to experimentally measure heat changes that accompany chemical reactions.  Reactions carried out inside a calorimeter result with heat flowing either into or out of the calorimeter. Therefore, exothermic reactions cause the temperature of the calorimeter to increase, while endothermic reactions result in a decrease in the temperature of the calorimeter.

Polystyrene Cup Calorimeters
For reactions that take place in aqueous medium, a simple, yet effective, calorimeter can be made from polystyrene cups.  In a polystyrene cup calorimeter, the reaction solutions are mixed and the heat associated with the reaction causes a change in temperature of both the contents in the cup and the cup itself.

The amount of heat associated with the reaction (q_rxn) will have a magnitude equal to but opposite in sign to the heat change of the polystyrene cup calorimeter (q_calorimeter). This relationship can be expressed as:

   -q_rxn = q_calorimeter

As already stated, the calorimeter is composed of both a solution in which the reaction takes place and the cup which contains the reaction solution. The heat of the reaction effects a heat change for both the solution (q_solution) and the cup (q_cup).  Therefore, the heat change of the calorimeter (q_calorimeter) is equal to the sum of the heat change of the solution (q_solution) and the heat change of the cup (q_cup).

    q_calorimeter = q_solution + q_cup

As previously indicated, the heat change of the reaction solution can be calculated from its mass, specific heat, and
temperature change.

    q_solution = (mass of solution)(specific heat of solution) (DT)

Similarly, the heat change of the cup can be calculated from the heat capacity of the cup and the temperature change.

    q_cup = (heat capacity of the cup)(DT)

Recall that in an acid-base neutralization that: Acid + Base -----> Salt + Water + Heat

The theoretical value for the heat liberated when a strong acid and strong base are neutralized is -56.8 kJ per mole of water formed.

EXPERIMENTAL PROCEDURES

I.  Measurement of the Heat Capacity of a Polystyrene Cup Calorimeter

1.  Obtain two thermometers and note any difference in the two measured temperatures.  The two thermometers should "agree."  If there are differences, this must be accounted for in all subsequent measurements.

2.  Using a graduated cylinder, place 50.0 mL of tap water in a clean, dry, polystyrene cup and place a thermometer in it supported by a ring and ring stand.  Place the lid on the cup and allow the water and cup to sit and equilibrate to a constant temperature.  Record the temperature of the cup and water as the room temperature (T_R).

3.  In another polystyrene cup, place 50.0 mL of hot water. The hot water can be obtained from a coffee pot.  Place a lid on the cup and place a thermometer in the hot water.  Support the thermometer by the ring and ring stand.

4.  Shortly after the lid has been placed on the hot water cup, begin timing and measure the temperature in the two cups at one-minute intervals for three minutes.  Record these measurements in the table on the data sheet.

5.  At the end of three minutes, quickly remove the lids and pour all of the water from the warm cup into the other cup.  Replace the lid and record the temperature of the mixture at one-minute intervals for three minutes.

         (Please note that the total timing period for steps 4 and 5, will be seven minutes.)

6.  Plot a single graph of  ^oC vs. time, which includes the data for the warmer water, cooler water and the water mixture.  Extrapolate the temperatures of the warmer water, the cooler water and the water mixture to the time of mixing.  Calculate the heat capacity of the cup.
 

II.  Measurement of the Heat of Reaction of a Strong Acid and a Strong Base

1.  Using two separate graduated cylinders, place 50.0 mL of 1.0 M HCl in a clean, dry polystyrene cup and 50.0 mL of 1.0 M NaOH in another clean, dry cup.  Place lids on the cups and measure and record the temperatures of the two solutions at one-minute intervals for three minutes.

2.  At exactly the four minute mark, remove the lids, and quickly but carefully quickly pour the HCl solution into the NaOH solution.  Replace the lid on the reaction mixture and continue recording the temperature of the reaction mixture at one-minute intervals for an additional three minutes.  Take care when removing and replacing the lids!

         (Please note that the total timing period for steps 1 and 2, will be seven minutes.)

3.  On a separate sheet of graph paper, plot a single graph of  ?C vs. time, which includes the data for the acid solution, the base solution, and the reaction mixture.  Extrapolate the temperatures of all three of the solutions (HCl, NaOH, and reaction mixture) to the time of mixing.  Using the data, calculate the heat of neutralization for the reaction of HCl(aq) and NaOH(aq).
 

DATA AND CALCULATIONS

I. Measurement of the Heat Capacity of a Polystyrene Cup Calorimeter

A.  Room Temperature (T_R)= ________ °C

B.  Temperature of water

	1 min.	2 min.	3 min.	4 min. (mixing)	5 min.	6 min.	7 min.
Warmer water (°C)
Cooler  water (°C)
Mixture (°C)

Temperature at time of mixing by extrapolation from graph:

        Warmer water (T_H) = ________ °C

        Cooler water (T_L) = ________ °C

        Mixture (T_M) = ________ °C

C.  Heat change of the warmer water =  (mass of water) x (specific heat) x (DT_W) = ________J
 
 

D.  Heat change of the cooler water = (mass of water) x (specific heat) x (DT_C) = ________J
 
 

E.  Heat lost to cup =  -("C"+"D") =________J
 
 

F.  Heat capacity of cup =  ("E"/DT_MR) = ________J/° C
 

DT calculations: DTW = TM - TH;  DTC = TM - TL;  DTMR = TM - TR density H2O = 1.0 g/mL;  specific heat H2O = 4.18 J/g° C

A.  Temperature of solutions and reaction mixture

	1 min.	2 min.	3 min.	4 min. (mixing)	5 min.	6 min.	7 min.
HCl (oC)
NaOH (oC)
Mixture (oC)

Temperature at time of mixing by extrapolation from graph:

            HCl (T_Acid) = ________°C

            NaOH (T_Base) = ________ °C

            Mixture (T_M)  = ________°C

           DT = T_M – [(T_Acid + T_Base )/2] = ________°C

B.  Heat change of the reaction mixture = [(mL of solution) x (density)] x (specific heat) x (DT) = ________J
 
 

C.  Heat change of the cup = (heat capacity of cup) x (DT) = ________J
 
 

D.  Total heat of reaction = -("B" + "C") = ________J
 
 

E.  Heat of neutralization per mole of water formed = "D"/moles H₂O = ________J/mole

                                                                                                                = ________kJ/mol

                HCl + NaOH ----> NaCl + H₂O
 

density of reaction solution, 0.5 M NaCl = 1.02 g/mL specific heat of reaction solution, 0.5M NaCl = 4.01 J/g° C mol of H2O = mol of HCl or mol of NaOH reacted = (L of HCl)(Molarity of HCl)