Activity: Heat and Temperature of Water Vapor (Remote)

In this introduction to heat capacity, students determine a derivative that indicates how much the internal energy changes as the temperature changes when volume is held constant.
What students learn
  • Heat capacity at constant volume relates to changes in internal energy, i.e. \({\mathit{\unicode{273}}} Q = dU\).
  • A conceptual definition of heat capacity at constant volume is the derivative of internal energy with respect to temperature without changing the volume.
  • Not all derivatives are slopes but they are all ratios of small changes.
  • Heat capacity at constant volume depends on the value of the volume. optional: Heat capacity at constant pressure is NOT \(\left(dU/dT\right)_p\) - you have to account for work done.
  • Media
    • activity_media/heat_capacity_tpu_sv_horizlabels_axis.pdf
    • activity_media/heat_capacity_tpu_sv_horizlabels_axis.png

Goals

  • Heat capacity at constant volume relates to changes in internal energy, i.e. \({\mathit{\unicode{273}}} Q = dU\).
  • A conceptual definition of heat capacity at constant volume is the derivative of internal energy with respect to temperature without changing the volume.
  • Not all derivatives are slopes but they are all ratios of small changes.
  • Heat capacity at constant volume depends on the value of the volume. optional: Heat capacity at constant pressure is NOT \(\left(dU/dT\right)_p\) - you have to account for work done.

Equipment Needed:

  • Heat Capacity graph
  • Optional: Blue \(U(T,p)\) plastic surface for each group
  • Student worksheet for each student
  • A personal or shared writing space for each student to write/draw/sketch.

Introduction

  • Students should have seen 1st Law of Thermodynamics and the Thermodynamic Identity
  • Students should know how to compute work as \(p\;dV\)

Whole Class Discussion:

  • Students need to figure out what derivative they're trying to estimate. Having an early WCD about this would be useful.
  • The derivative they are trying to estimate is tantalizingly close to a slope, especially for the blue dot and the green triangle. Not all partial derivatives are slopes but they are all ratios of small changes.
  • For an ideal gas, the heat capacities are constant. Water vapor is not an ideal gas but it is pretty close. In general, the heat capacity is a state variable.

Heat & Temperature of Water Vapor

A pressure cooker is an enclosed pot that expels air and traps water vapor, which increases the internal pressure. This in turn raises the boiling point of water and allows food to cook at high temperatures.


Imagine you have a large industrial pressure cooker that holds 1 kg of water vapor. You would like to know how responsive the system is to changes in temperature. To do this, you need to determine a characteristic rate: how much heat is needed to change the temperature by a small amount.


The graph shows internal energy and volume contours plotted on temperature and pressure axes.


Internal Energy 2cm \(\rightarrow\) 170. kJ
Temperature 2cm \(\rightarrow\) 70 K
Pressure 2cm \(\rightarrow\) 128000 Pa
Entropy Contours Curves \(\rightarrow\) 0.33 kJ/K apart
Volume Contours Line Segments \(\rightarrow\) 0.7 m3 apart


Estimate: Use the graph to determine this temperature-responsiveness when the volume is held fixed. The initial state of the system corresponds to the black square. Describe your process.

Student Ideas: Students tend toward trying to use equipartition theorem or other equations, rather than directly measuring the rate on the surface. Students might not always realize that work done is zero, or how this would relate U and Q.

Solution Want the ratio of change in internal energy and the change in temperature. This is NOT the slope of the surface. It is the ratio of small changes.

Discussion: We haven't told students what derivative to estimate. Have a whole class discussion about this after most groups have struggled for a few minutes and have had some insights.

Student Ideas: Students might need help in recognizing that for a pressure cooker, you can control the pressure and temperature but the volume of water vapor is constant.

Follow-Up: For what graph would the slope be the rate you're looking for? How is that graph related to the surface?


Explain: Why does it matter that you are holding volume constant in the above estimate?

Two reasons: (1) you need to specify a path in order to take the derivative, and (2) holding the volume constant means the work done is zero, so the heat added is equal to the change in internal energy.


Explore: Does the value of your estimate depend on the value of the volume?

Any answer is productive here. For an ideal gas, the answer is no. Water vapor is nearly an ideal gas (but not quite). The heat capacity at constant volume is not the directional derivative.

Optional Extensions:
  • Can you estimate the heat capacity at constant PRESSURE? (Yes, there are several ways, including estimating the work done along the constant pressure path and subtracting that from the change in internal energy. Also, chain rule (try to avoid).)

  • Can you estimate \(C_v\) using a different graph?

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    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.
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    6. 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.

Learning Outcomes