H-3  Latent heat of vaporization of liquid-N$_2$

OBJECTIVE:

To measure the heat of vaporization of liquid nitrogen, L$_{v}$, at its boiling point (T$_{b}$ = 77 K at standard atmospheric pressure).

APPARATUS:

Dewar flask; liquid nitrogen (ask for help in getting it from a large storage Dewar opposite room 4411 Sterling); aluminum cylinder on a long thread; double pan balance; calorimeter plus water jacket for thermal ballast (as in H-2a); timer; thermocouple type digital thermometer; selection of slotted masses; coffee pot for hot water.

PRECAUTIONS:

Liquid N$_2$ is fascinating to work with. However, keep in mind the following simple safety precautions.

1.

Never stopper a flask of liquid N$_2$ with an unperforated stopper.

2.

Have a perforated stopper on the Dewar throughout the experiment to prevent condensation of moisture from the air on the inside of the flask.

3.

Avoid prolonged contact of liquid N$_2$ with your skin. The insulating vapor layer may disappear and severe frost-bite may result.

INTRODUCTION:

When one lowers an aluminum cylinder of mass, m$_{Al}$, and at room temperature, T$_{r}$, into liquid N$_2$ at its boiling temperature, T$_{b}$, the cylinder cools to T$_{b}$. The heat given off during this cooling, Q$_{Al}$, will vaporize a mass m$_{N}$ of liquid N$_2$.

You might expect to find $L_{v}$, of the nitrogen by setting

$$m_{N}L_{v} = Q_{Al} = m_{Al}c_{Al}(T_{r} - T_{b})   .$$

This method fails because $c_{Al}$ is not constant over the $\sim 220^o\mbox{C}$ temperature range between T${_r}$ and T${_b}$. See figure 1.

We can avoid this difficulty by noting that Q$_{Al}$ is also the heat needed to warm the same aluminum cylinder to from T$_{b}$ to T$_{r}$. You can measure this heat by placing the cold aluminum cylinder (at temperature $T_{b}$) in a ``calorimeter'' that contains water and observing the change in temperature of the water, $-\Delta T$, -provided that the final temperature of the water, $T_{f}$, is room temperature, $T_{r}$. It is hard to arrange for $T_{f}$ to end up exactly at room temperature, but if $T_{f}$ is close to $T_{r}$, one can accurately correct the calorimeter data for the small additional heat term, namely $m_{Al}c_{Al}(T_{r} - T_{f}$), since over the small $T_{r} -T_{f}$ interval, $c_{Al}=0.212  \mbox{cal/g/K}$ is constant.

Figure 1: Specific heat capacity of Al vs. temperature.

SUGGESTIONS ON PROCEDURE:

1.

To maximize sensitivity (i.e. to get a large temperature change in the water) use only enough water ($\sim$125-150 grams) in the calorimeter to cover the metal cylinder.

2.

The calorimeter is designed to thermally isolate the water from the surroundings. The inner vessel of the calorimeter is mounted within, but thermally isolated from, a surrounding water jacket, which is close to room temperature. The isolation isn't perfect, so here will be some small amount of heat flow between the calorimeter and the jacket. To minimize the net heat exchange with the water jacket, you will want to make the initial water temperature as far above the jacket temperature as you expect it to end up below the jacket temperature after the water in the inner vessel has been cooled by your Al cylinder. (The jacket temperature will remain fairly constant during the experiment.) Estimate roughly the proper initial water and calorimeter temperature. [Use the measured mass of the aluminum cylinder, m$_{Al}$, its specific heat (0.212 kcal/kg$^o$C) and the b.p. of liquid N$_2$, $T_{b} = 77 \mbox{K}$. For this rough calculation, assume that the specific heat of Al $\it is$ constant with temperature.] Determine the water mass, $m_{w}$, and appropriate starting temperature.

3.

Place the flask containing liquid nitrogen (plus perforated stopper) on one pan of a balance and record the mass each minute for 10 minutes. (Why is the mass decreasing?)

4.

Record the temperature of the metal cylinder, T$_{R}$, and then lower it (by a thread) gently to the bottom of the flask. Replace the stopper (perforated) on the top of the flask and continue recording the total mass each minute until it shows a slow steady decrease.

5.

Record the initial temperature, $T_{i}$, of the calorimeter, which you have chosen so cleverly in section 2. Transfer the cold metal cylinder into the calorimeter, and note the calorimeter temperature every two minutes (while gently stirring). Record the mass of the flask of liquid N$_2$ on alternate minutes. When the calorimeter temperature has reached a slow steady rate of change and the mass of the flask of liquid N$_2$ is falling at a slow steady rate, discontinue the readings.

6.

Plot the mass of the flask plus nitrogen as a function of time. For the minutes that the cylinder of metal was in the flask, subtract the mass of the cylinder. See figure for a typical plot.

How much of the N$_2$ mass change was caused by the heat from the cylinder? If a-b and c-d were parallel, it would be the vertical distance between these lines. But c-d ordinarily has a smaller slope than a-b, possibly because the evaporation of liquid nitrogen between b and c has cooled the upper part of the flask.

Hence we use the average of the two rates of fall by drawing a vertical line through e, (the midpoint of line b-c). Then f-g estimates the mass, m$_{N}$, evaporated by the heat from the cylinder.

7.

The final temperature, $T_{f}$, of the Al cylinder in the calorimeter usually will not be quite the same as the initial temperature ($= T_{r}$) of the cylinder before it was lowered into the liquid nitrogen.

Hence $Q_{Al}$ will be the heat to warm the Al cylinder in the calorimeter to $T_{f}$ plus the mass of the cylinder $\times$ (specific heat of aluminum) $\times (T_r - T_f)$.

Specifically if:

         $L_{v}$ = latent heat of vaporization of nitrogen

  $m_{N}$ = mass of nitrogen evaporated by heat from the cylinder

  $m_{Al}$, c$_{Al}$ = mass and specific heat of Al cylinder

         $T_{i}$ = initial temperature of water and calorimeter

  $m_{w}$,  c$_{w}$  		 = mass and specific heat of water

  $m_{c}$,  c$_{c}$  		 = mass and specific heat of calorimeter and stirrer

         h$_{t}$ = heat capacity of immersed part of thermometer,

then, if we neglect $h_{t}$:

$$Q_{Al} = m_N L_v = (m_w c_w + m_c c_c) (T_i - T_f)  + m_{Al} c_{Al} (T_r - T_f)  .$$

Calculate $L_v$ from the above relation. The accepted value is 47.8 kcal/kg.

OPTIONAL:

1. Calculate the apparent specific heat of the Al block by use of
   
   $$c_{Al} = \frac{Q}{m_{Al}(T_r - T_b)}$$
   
   
   and your data. How does your result compare with the accepted value of $c_{Al}$ = 0.212 kcal/kg? Explain. (See Introduction).
2. Observe (but do not touch) the following items after immersion in liquid N$_2$: rubber (get a piece from the instructor), pencil eraser.
3. Pour a little liquid N$_2$ onto the floor. Explain the behavior of the small spheres of liquid N$_2$.