4.1.2: Explorations

Exploration 1: Mechanical Equivalent of Heat

As the \(100\text{-kg}\) red mass drops, the paddle turns in the liquid and the liquid heats up. Joule used a version of this device to determine the equivalence between heat and work. You will run the animation to do the same (position is given in meters, time is given in seconds, and temperature is given in degrees Celsius). The temperature of the liquid is given by the thermometer shown. Restart.

The dimension of the container that holds the blue liquid that you cannot see (into the screen) is \(0.1\text{ m}\). The density of the liquid is \(13,600\text{ kg/m}^{3}\).

1. What is the volume of the liquid?
2. What is the mass of the liquid?
3. During the animation, what is the change in temperature of the liquid?
4. If it takes \(33\) calories to raise \(1\text{ kg}\) of the liquid \(1^{\circ}\text{C}\), how much heat goes into the liquid?
5. What is the change in kinetic energy of the falling red mass?
6. What is the work done by gravity on the mass (in joules)?
7. The work in (f) goes into frictional heating of the liquid (as the paddles turn through the liquid). Therefore, how many calories are equal to \(1\) Joule?

Exploration authored by Anne J. Cox.
Script authored by Anne J. Cox and Mario Belloni.

Exploration 2: Expansion of Materials

A rod is fixed at one end. In the animation you see both the rod and a magnified view of the right end (position is given in meters, time is given in minutes, and temperature is given in kelvin). As you increase the temperature, notice that the rod increases in length. This Exploration will help you develop a quantitative relationship for the increase in the length of the rod, as a function of the initial length and the temperature change, that holds for all materials. Restart.

Note that the \(x10\) closeup means that a reading (in meters) really is in tenths of meters.

1. For Animation 1, if you double the length, what happens to the change in length?
2. Repeat (a) for the material in Animation 2. How do the two results compare?
3. How does changing the final temperature change the expansion? (If you double the change in temperature, what happens to the change in length?)
4. What general expression can you now write for the change in length as a function of the temperature change and initial length?

The difference between the two materials is described by a different coefficient of linear expansion, \(\alpha\). For the material in Animation 1, \(\alpha\) is \(30\times 10^{-6}\text{/K}\), while for the material in Animation 2, \(\alpha\) is \(20\times 10^{-6}\text{/K}\).

When heated, a solid (even a thin rod as above), expands in all three dimensions. The equation for the volume expansion is similar to the linear expansion case with the coefficient of expansion approximately equal to \(3\alpha\).

5. Why didn't you see the expansion of the rod in the other dimensions?

Exploration authored by Anne J. Cox.

Exploration 3: Calorimetry

When two objects at different temperatures are in thermal contact with each other, they will eventually reach the same temperature. Heat will flow from the warmer object to the cooler one until they are at the same temperature (temperature is given in kelvin and heat is given in joules). Using the equation for heat absorbed or released with a temperature change, \(Q = mc (T_{f} - T_{i})\), where \(Q\) is the heat, \(m\) is the mass, \(c\) is the specific heat, and \(T\) is the temperature (with subscripts indicating final and initial temperatures), we can determine the specific heat of an object. Restart.

In the animation a piece of metal at a high temperature is placed in water at a lower temperature and, if the water is well insulated so we can assume essentially no heat loss to the environment, the final temperature of the water/metal combination depends on the mass of the water, the mass of the metal, and the specific heat of each. Try changing the initial temperature of the metal and the mass of the metal. By equating the heat lost by the object to the heat gained by the water, we can calculate the specific heat of an unknown object or the final temperature of the system. There are \(10\text{ kg}\) of water, and the specific heat of water is \(4.186\text{ kJ/kg}\cdot\text{K}\). The specific heat of the block is \(0.39\text{ kJ/kg}\cdot\text{K}\).

1. For a \(1\text{-kg}\) block and an initial block temperature of \(800\text{ K}\), use the equation above to calculate the heat absorbed by the water and the heat released by the block when they reach the final temperature.
2. What is the scale for the heat on the bar graphs? In other words, what unit of heat does each mark correspond to (\(10\text{ kJ},\: 100\text{ kJ},\: 200\text{ kJ}\), etc.)?
3. If \(m = 3\text{ kg}\) and the initial temperature of the block is \(1000\text{ K}\), equate the heat released to the heat absorbed to predict the final temperature. Run the animation to check your prediction of both the final temperature and the heat released and absorbed.

Exploration authored by Anne J. Cox.

Exploration 4: Heat Balance

The animation depicts an incubator for chicks. Inside the box is a heat lamp with varying power. The incubator is made of material with a thermal conductivity of \(0.15\text{ W/m}\cdot\text{K}\) and of \(2\text{-cm}\) thickness. The dimension of the box you cannot see (the depth of the box going into the screen) is \(0.3\text{ m}\) (position is given in tenths of meters and temperature is given in degrees Celsius). As you change the power of the heat lamp, notice how the inside equilibrium temperature changes, as does the energy/time (power) lost by conduction through the walls to the outside. Restart.

The box must also be coated in a shiny reflective material (foil) so that there are no significant contributions through radiative heat-exchange processes. The only significant energy-exchange process is conductive.

1. The animation shows an instantaneous change in temperature as you change the heater power. Explain why this is nonphysical (the temperature does not change instantly as shown). What determines how long it will actually take for the system to reach equilibrium?
2. When the heater runs at \(50\text{ W}\), calculate the energy lost through conduction using \(P = (kA/x)\Delta T\), where \(k\) is the thermal conductivity (\(0.15\text{ W/m}\cdot\text{K}\) in this case), \(A\) is the surface area of the box, \(x\) is the thickness of the box material, and \(\Delta T\) is the temperature difference between the inside of the incubator and the outside environment. This should be equal to the power delivered by the heater (\(50\text{ W}\)).
3. By equating the energy/time into the box (from the heat lamp) to the energy lost from the box (by conduction out through the walls), predict the power required from the heat lamp to keep the box at \(27^{\circ}\text{C}\).
4. Check your prediction by varying the heat lamp power. Note that if there are chicks in the incubator, they will radiate heat as well, and so the heat required from the heat lamp will be reduced.

Exploration authored by Anne J. Cox.

Physlets were developed at Davidson College and converted from Java to JavaScript using the SwingJS system developed at St. Olaf College.