1. CALIBRATION - DETERMINATION OF THE CALORIMETRIC CONSTANT

How to calculate T when two samples of water of known mass and temperature are mixed.

E.g., suppose

60 g of hot water at 50^oC, are added to 30 g of cold water at 25^oC in a calorimeter

Ideally,

heat gain by cold water         =                 -  heat lost by hot water

Since heat gained or absorbed is positive while heat lost is negative, we must put a negative sign on the heat lost for the equality to hold.

q_cold    =      m_c C_c ΔT_c          =         - q_hot          =      - m_h C_h ΔT_h

30 g * 4.18 J/g * ( T_mix - 25 )     =           - 60 g * 4.18 J/g * (T_mix - 50 )

30 * ( T_mix - 25 )       =            - 60 * ( T_mix - 50 )

                               T_mix           =          ( 60 * 50 + 30 * 25 ) / ( 30 + 60 )

         =         42^oC

The actual  T_mix will differ from this value because of some heat loss to the container - i.e., calorimeter

Assume the calorimeter experiences the same temperature change as the cold water. (The cold water was originally in, and at the same temperature as, the calorimeter.) This is determined experimentally by measuring the temperatures of the relevant water samples before and after mixing. (Example of such a determination.)

The heat lost by the hot water now goes to increasing temperature of both the cold water and the calorimeter.

Now we have

-q_hot        =          q_cold + q_cal

or,          q_cal = - q_hot - q_cold

Suppose, for example, that the actual (measured) final temperature of the above mixture is 40^oC

-q_hot  = - 60 g * 4.18 J/g * ( T_mix - 50) = 2510 J

q_cold =   30 g * 4.18 J/g * ( T_mix - 25) = 1880 J

Note that since some heat must be transferred to the calorimeter, the magnitude of the heat lost by the hot water must be greater than the magnitude of the heat gained by the cold water.

q_cal = -q_hot - q_cold = 2510 - 1880 = 630 J

ΔT for the calorimeter = ΔT for the cold water = 15 C^o

It behaves like an object with a heat capacity

C_cal = q_cal / ΔT = 630 / 15 = 42 J / C^o

2. DETERMINATION OF ΔH_neu FOR HCl

It is important to keep track of which quantitites are measured in the exercise and which ones are given.

Suppose we mix samples of measured volumes and temperatures and given concentrations of HCl and NaOH

Measure temperature change at t = 0 graphically. (Here is the graph.) The result is that ΔT = 4.75C^o

Using the measured volumes and given concentrations of HCl and NaOH, we can calculate the number of moles of each reactant.

n_HCl     = V_HCl  * M_HCl = 0.0600 mol
      n_NaOH = V_NaOH * M_NaOH = 0.0715 mol

HCl is the limiting reagent. We can neutralize only 0.0600 mol HCl

Using the given heat capacity, densities and measured volumes of the solutions and the measured temperature change, we can analyze the heat exchange, q, and correct for the heat loss to the calorimeter using the Calorimetric constant, C_cal, determined in Part 1.

Heat Capacity = 3.97 J/g
Density = 1.02 g/mL
C_cal = 42.0 J/C^o

Tot Vol of Sol = 50.0 + 55.0 = 105.0 mL

From the given density, we can calculate the total weight of this volume of solution

Tot Wt Sol = 1.02 * 105.0 = 107 g

From the given heat capacity of the solution and the measured temperature change, we can calculate the amout of heat absorbed by the solution.

q_SOL = 3.97 J/gC^o * 107 g * 4.75C^o = 2020 J (actually, only 3 significant figures are justified)

But, some of the liberated heat went to raising the temperature of the calorimeter. Using the calorimetric constant and the measured temperature change, we can calculate this additional amount of heat.

q_CAL = 42.0 J/C^o * 4.75C^o = 200 J

The total number of Joules liberated by the reaction is then:

q_RXN = 2020 J + 200 J = 2220 J

But the reaction involved the neutralization of 0.0600 mol of HCl. therefore, the molar heat of neutralization was

q_RXN = 2200 J / 0.0600 mol = 3.67 X 10⁴ J/mol

But the change in enthalpy of a process is equal to the heat gained (q) by the system. Since the reaction liberated heat rather than gaining it, the heat, and therefore, the enthalpy of the neutralization reaction is the negative of the above amount, or

ΔH_neut for HCl = - 3.67 X 10⁴ J/mol