Student handout: Ice Calorimetry Lab

This lab gives students a chance to take data on the first day of class (or later, but I prefer to do it the first day of class). It provides an immediate context for thermodynamics, and also gives them a chance to experimentally measure a change in entropy. Students are required to measure the energy required to melt ice and raise the temperature of water, and measure the change in entropy by integrating the heat capacity.
What students learn This lab gives students a chance to take data on the first day of class (or later, but I prefer to do it the first day of class). It provides an immediate context for thermodynamics, and also gives them a chance to experimentally measure a change in entropy. Students are required to measure the energy required to melt ice and raise the temperature of water, and measure the change in entropy by integrating the heat capacity.
In this lab, we will be measuring how much energy it takes to melt ice and heat water.
Materials:
  • Styrofoam cup
  • Heating element
  • Scale
  • 2 digital multimeters
  • Temperature guage
  • Ice and water

The setup

You will put some mass of ice (about 50g) and ice-cold water (about 150g) into your styrofoam cup. Use the scale to record the mass of the ice and water as you add them to the cup. Finally, add your ice-cold heating element and thermometer through the lid of the cup.

Collect data

We will be measuring the temperature of the water and the power dissipated in the heating element (which is just a resistor). Thus we can find out how much energy was added to the water, and how this changes the temperature. In order to keep the temperature measurement reasonable, we will need to periodically stir the cup and heat it moderately slowly.

You will be collecting temperature data using the computer, so before you turn on the heater, you should make sure the computer is taking data. Turn on the heater, and write down the time you do so as well as the current and voltage, from which you can find the power dissipated in the resistor. If the current or voltage changes during the course of the experiment, take note of the new values---and the time.

Question 1: Plot your data I
Plot the temperature versus total energy added to the system (which you can call \(Q\)). To do this, you will need to integrate the power. Discuss this curve and any interesting features you notice on it.

Question 2: Plot your data II
Plot the heat capacity versus temperature. This will be a bit trickier. You can find the heat capacity from the previous plot by looking at the slope. \begin{align} C_p &= \left(\frac{\partial Q}{\partial T}\right)_p \end{align} This is what is called the heat capacity, which is the amount of energy needed to change the temperature by a given amount. The \(p\) subscript means that your measurement was made at constant pressure. This heat capacity is actually the total heat capacity of everything you put in the calorimeter, which includes the resistor and thermometer.

Question 3: Specific heat
From your plot of \(C_p(T)\), work out the heat capacity per unit mass of water. You may assume the effect of the resistor and thermometer are negligible. How does your answer compare with the prediction of the Dulong-Petit law?

Question 4: Latent heat of fusion
  1. What did the temperature do while the ice was melting? How much energy was required to melt the ice in your calorimeter? How much energy was required per unit mass? per molecule?
  2. The change in entropy is easy to measure for a reversible isothermal process (such as the slow melting of ice), it is just \begin{align} \Delta S &= \frac{Q}{T} \end{align} where \(Q\) is the energy thermally added to the system and \(T\) is the temperature in Kelvin. What is was change in the entropy of the ice you melted? What was the change in entropy per molecule? What was the change in entropy per molecule divided by Boltzmann's constant?

Question 5: Entropy for a temperature change
Choose two temperatures that your water reached (after the ice melted), and find the change in the entropy of your water. This change is given by \begin{align} \Delta S &= \int \frac{{\mathit{\unicode{273}}} Q}{T} \\ &= \int_{t_i}^{t_f} \frac{P(t)}{T(t)}dt \end{align} where \(P(t)\) is the heater power as a function of time and \(T(t)\) is the temperature, also as a function of time.

Author Information
David Roundy
Keywords
heat entropy water ice
Learning Outcomes