Calorimetry question

Hey guys.

Could anyone tell me all the steps to do an accurate calorimetry experiment, including any graphing, extrapolating etc.
And give reason for steps such as seting temp of water equal to solution if using, and the reason for extrapolating etc.

Many Thanks

Reply 1

marcusfox

Basically, you need to think about how accurate you're going to want to be, and whether you're going to need an expensive, highly accurate, computer controlled, specialist calorimeter or just use one that the laboratory has available, which will probably be something just slightly more advanced than polystyrene coffee cups.

What do you mean by "seting[sic] temp of water equal to solution if using"? Surely you can directly measure the temperature change of a solution with a thermometer.

First you need to calibrate the calorimeter. I will assume you're using polystyrene coffee cups.

The ability of the calorimeter to absorb(or release) heat is called the heat capacity of the calorimeter (C_cal). It is determined through a simple experiment. The heat capcity will have units of energy/temperature (ex. J/°C) and can be determined as follows.

A known mass of cold water is placed in the coffee cup calorimeter and allowed to equilibrate in temeperature with the cups etc. A known mass of hot water is prepared and its temperature determined. The hot water is quickly added to the cold water in the calorimeter. At this point three things occur simultaneously:

the hot water is cooled
the cold water is warmed
the calorimeter is also warmed
All of the heat is trapped in the calorimeter (we assume none escapes to the surroundings) so the total heat conservation equation is:

(D=Delta, funny greek triangle thingy :confused:

)

0 = q_cal + q_{hot water} + q_{cold water}
q_cal = C_cal x DT_cal and
q_{cold water} = mass_{cold water} x DT_{cold water} x 4.184 J/°C g
q_{hot water} = mass_{hot water} x DT_{hot water} x 4.184 J/°C g
4.184 J/°C g is called the specific heat of water because 4.184 Joules is the amount of heat that must be added or removed from one gram of liquid water to change the temperature of one gram of water by one °C.

It is critical to keep track of the signs of all the variables. When finding DT always subtract the initial temperature from the final temperature. In this experiment, two of the DT terms will be equal and the third will differ from the others in both sign and magnitude.

Now you do whatever experiment you do.

The change in temperature is determined by measuring the initial temperature, T₁, and the maximum temperature, T₂, of the contents of the calorimeter during the experiment. The determination of a precise value for T₂ is complicated by the fact that a small heat exchange occurs between the surroundings and the contents of the calorimeter, both during the reaction and after its completion. The rate of exchange depends on the insulating properties of the calorimeter and on the rate of stirring. A correction for this heat loss is made by an extrapolation of a temperature vs. time curve. By extrapolating the linear portion of the curve back to zero time (the time of mixing of the reactants in the calorimeter), T₂ is obtained. The value of T₂ obtained in this manner is the temperature of the product solution, had the reaction occurred instantaneously with no enthalpy loss to the surroundings.

q_sys is the resulting measurement. Then you use 0 = q_cal + q_sys to get your result.

Marcus

Reply 2

slaw

Hm. Interesting. Well, I've always been taught to make sure the water remains the same temperature for like one minute or something. Anyway, the real question is should you extrapolate or take readings over a period of time for a calorimeter made from polystrene cups. Because my current teacher says that there will be insignificant heat lost so it wouldn't change the calculations too much but my old teacher taught us to extrapolate the readings to get the "corrected temp rise"
And i like your
0 = qcal + qhot water + qcold water

Reply 3

junners

Can't you do both?

I've always been told to measure the temperature for the 1st, 2nd and 3rd minute.

Then mix the 2 substances at the 4th minute.

Then measure the temperature at minute-intervals, from the 5th minute, until you get 2 concordant results.

Then plot a graph of temperature vs. time and extrapolate both lines (i.e. the line before the 4th minute and the line after the 4th minute).

Therefore the change in temperature is the difference between the 2 extrapolated lines at the 4th minute.

I've been taught that this is one way of reducing the errors from the heat loss due to the apparatus.