1. Mixing Experiments

In this experiment we will build on the relationship,

Q = cmDT,

(1)

where Q is the quantity of heat transferred to a system, m is the mass of the system, and DT is its temperature change. c is a constant characteristic of the substance. (Since water will be the substance in all of the work today, we don't have to have a value for c; it will always cancel.)

Equation (1) applies whether the temperature is increasing or decreasing. If the temperature is increasing, the process is a flow of heat into the water; if the temperature is decreasing, the process is a flow of heat out of the water.

Suppose we mix together hot water (temperature T_hot) and cold water (temperature T_cold). The water eventually comes to an equilibrium at a final temperature, T_final, between T_hot and T_cold. The conservation law to be investigated here is, does the heat that flows out of the hot body equal the heat that flows into the cold body?

The heat that enters the cold water is

Qcold = cmcold(Tfinal - Tcold).

(2)

The heat that leaves the hot water is

Qhot = cmhot(Thot - Tfinal).

(3)

m_hot and m_cold are the masses.

Place 150 ml of water in one beaker and the same volume of water in a second beaker. Heat one beaker to a temperature between 70 and 80 degrees. Record the temperatures of both, and then combine them in the styrofoam cup. Mix gently for a few seconds, and then read the final temperature.

Given that the two quantities of water have the same volume -- and same mass -- what do you expect the final temperature to be? How close does your measurement come to this value? Calculate the quantities Q_cold and Q_hot from (2) and (3) above, leaving c as a coefficient). How close are they? Find the difference as a percentage.

Now repeat the experiment using 200 ml of cold water and 100 ml of hot water. Looking at your values of T_hot and T_cold, what would be your estimate of the final temperature after mixing? Again calculate Q_cold and Q_hot from Eqs. (2) and (3), and find the percent difference.

If there seems to be time (allowing 45 minutes to an hour for part 2 below), try the experiment with 200 ml of hot water and 100 ml of cold water.

2. Change of Phase

Eq. (1), saying that temperature increases when heat is added to a system, fails dramatically when the system reaches a temperature where it changes phase. Boiling water and melting ice are the most familiar examples. When the temperature of water reaches the boiling point, continued heating does not raise the temperature. It stays at 100 degrees as the water changes to steam. The heat needed to boil a given quantity of liquid is proportional to the mass that boils:

Q = mL,

(4)

where L is called the latent heat of boiling.

In this experiment we will heat water from room temperature to boiling, measuring the time and the temperature, and then let it boil for a known time. We will measure the mass that boils, and then compare the heat needed for raising temperature to the heat needed for boiling.

We will want a known mass of water, so weigh the empty beaker, and then weigh it with about 200 ml of water. Set up the bunsen burner as in the previous experiment, and let the position of the flame be unchanged during the experiment. Begin heating, start the clock, and record time and temperature every minute.

Note carefully and record the time when boiling starts. This will be when you begin to see small bubbles forming. (The thermometer may read a few degrees below 100^o C. at this point, possibly because of an uneven temperature distribution.) Continue heating for 8 or 10 minutes. Take the beaker off the flame (recording the time that you do), let it cool for 2 or 3 minutes, and weigh it (to find the mass of water that boiled).

From the balance readings that you have, calculate

• mass of water that underwent the temperature rise (m_rise)
• the mass of water that boiled away (m_boil)

Since the mass that underwent the temperature rise was not the same as the mass that boiled, we will want to compare boiling to temperature change by calculating the following ratio:

heat needed to boil 1 gram of water	(5)
heat needed to raise temp. of 1 gram of water by 1 degree

Given the assumption that heat enters the water at a constant rate, the ratio of heats is equal to the ratio of times:

time to boil 1 gram of water	(6)
time to raise temp. of 1 gram of water by 1 degree

From your data, calculate the numerator and denominator of this ratio, and divide. The accepted value of this ratio is 540.

The denominator in ratio (5), the heat needed to raise the temperature of 1 g of water by 1 degree, is defined as the calorie. Hence the ratio you have determined is the latent heat of water, measured in calories per gram.

3. Latent Heat and Energy Since supplying heat can change the phase of a system, this change represents an increase in the internal energy of the system. Just as the molecules in 50^o water have on the average more energy than the molecules in 20^o water, the molecules in steam at 100^o have on the average more energy than the molecules in liquid water at at 100^o. In the case of the liquid, the molecules are close to each other, and subject to attractive forces. This means they have less energy than when they are separated (as in steam or other gases) and fly around in space with lots of kinetic energy.