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Lab 1: Energy and States of Matter

Part A: Heating Curve for Water

The first part of the lab is straightforward; prepare heating curves by plotting temperture (y-axis) vs time (x-axis) for each trial.  Determine the boiling point of the water from each plot, report and average value, and calculate the percentage error from the literature value.

Part C: Calorimetric Determination of the Heat of Fusion

Many processes dealing with energy changes and their corresponding temperature changes can be studied in an apparatus called a calorimeter.  A calorimeter is a well-insulated container that minimizes energy transfer between its contents (the system being studied) and the surroundings.  A reasonable calorimeter can actually be constructed from a couple styrofoam cups, since styrofoam has a very low coefficient of heat transfer.

For the purposes of this experiment we will assume that no energy is exchanged with the surroundings (it might be a good idea to consider how valid this assumption is in the discussion).  Therefore, if some process within the calorimeter loses energy, there must be some other process in the calorimeter that gains that energy.  Mathematically, that means the sum of all energy changes in the calorimeter must equal zero.

For the melting of the ice, we have several processes to consider:

ice (0 °C) → water (0 °C) → water (at final temperature Tf)
water (at initial temperature Ti) → water (at final temperature Tf)

In the first process, the ice in the calorimeter melts, and then the newly formed water warms to the final temperature.  In the second process, the water initially in the calorimeter is cooled to the final temperature.  Note that although the calorimeter is full of nothing but water at the end of the experiment, that water had two sources (the initial quantity of water and the ice).  We can represent the energies for these processes mathematically:

qmelting + qwarming + qcooling = 0

Substituting some specific variables into the equations we’ve discussed in class,

qmelt = mice · ΔHf
qwarming = mice · C · (Tf – 0 °C)
qcooling = mwater · C · (Tf – Ti)

In the lab, we determined mice and mwater by difference weighing the calorimeter, and we measured Ti and Tf.  The heat capacity of water is C = 4.184 J / g · °C, which means that if we combine the equations above, we can calculate ΔHf.  Note that qcooling will be negative because the cooling water is releasing energy.

For your results, calculate ΔHf from each trial and report and average value (with units of kJ / mol).  Using the literature value of ΔHf = 6.02 kJ / mol, calculate the percentage error.