Question 7.EC.2: Measurement of the Specific Latent Heat of Evaporation of Li......
Measurement of the Specific Latent Heat of Evaporation of Liquid Nitrogen
In this part you can use the following equipment:
(1) A Styrofoam cup with a cap.
(2) Dewar flask with liquid nitrogen.
(3) An aluminum cylinder with a hole (item 3, Part 1).
(4) Electronic scales with accuracy of 1 g (item 4, Part 1).
(5) A digital timer (item 5, Part 1).
(6) Plotting (graphing) paper (2 pages).
(7) Pieces of thread.
The specific latent heat of evaporation of water is well known, while rarely we have to deal with one the main atmospheric gases, nitrogen, in its liquid form. The boiling temperature of liquid nitrogen under normal atmospheric pressure is very low, T_{N} = 77 K =- 196°C.
In this experiment you are asked to measure the specific latent heat of evaporation of nitrogen. Because of heat exchange with the environment nitrogen in a Styrofoam cup evaporates and its mass decreases at some rate. When an aluminum cylinder initially at room temperature is immersed into nitrogen, nitrogen will boil violently until the temperature of the aluminum sample reaches the temperature of liquid nitrogen. The final brief ejection of some amount of vaporized nitrogen from the cup indicates that aluminum has stopped cooling – this ejection is caused by the disappearance of the vapor layer between aluminum and nitrogen. After aluminum reaches the temperature of nitrogen, the evaporation of nitrogen will continue.
When considering a wide range of temperatures, one can observe that the specific heat of aluminum C_{Al} depends on absolute temperature. The graph of aluminum’s specific heat in arbitrary units versus temperature is shown in Fig. 7 – 13. Use the result of specific heat measurement in 45°C-65°C temperature range in Part 1 to normalize this curve in absolute
Warning: (1) Liquid nitrogen has temperature T_{N}\,=-\,196^{\circ}\mathrm{C}. To prevent frostbite do not touch nitrogen or items which were in contact with nitrogen. Make sure to keep away your personal metal belongings such as jewelry, wrist watch, etc ..
(2) Do not put any irrelevant items into nitrogen.
(3) Be careful while putting the aluminum cylinder into the liquid nitrogen to prevent spurts or spilling.
The task
(2a) Measure the evaporation rate of nitrogen in a Styrofoam cup with a closed cap, and measure the mass of nitrogen evaporated during the cooling of the aluminum cylinder (aluminum cylinder is loaded through a hole in the cup). Proceed in the following manner. Set up the Styrofoam on the scales, pour about 250 g of liquid nitrogen in it, wait about 5 minutes and then start taking measurements. After some amount of nitrogen evaporates, immerse the aluminum cylinder into the cup – this will result in a violent boiling. After aluminum cylinder cools down to the temperature of nitrogen, evaporation calms down. You should continue taking measurements in this regime for about 5 minutes until some additional amount of nitrogen evaporates. During the whole process record the readings of the scales M(t) as a function of time.
IN NO CASE SHOULD YOU TOUCH THE ALUMINUM CYLINDER AFTER IT WAS SUBMERGED INTO LIQUID NITROGEN.
In your report provide a table of M(t) and m_{N}(t), where m_{N}(t) is the mass of the evaporated nitrogen.
(2b) Using the results of the measurements of M(t) in (2a), plot the graph of the mass of evaporated nitrogen m_{N} versus time t. The graph should illustrate all the three stages of the process – the calm periods before and after immersion of aluminum, and the violent boiling of nitrogen.
(2c) Determine from the graph the mass m_{N}^{{Al}} of nitrogen evaporated only due to heat exchange with the aluminum cylinder, as it is cooled down from room temperature to the temperature of liquid nitrogen. In order to do this you have to take into account the heat exchange with the environment through the cup before, during and after the cooling of aluminum. Write the value of m_{N}^{{Al}} and its uncertainty Δm_{N}^{{Al}} on the answer sheet.
(2d) Using the result of measurement of aluminum’s specific heat in the temperature range of 45°C – 65°C (Part 1), normalize the graph of the relationship between aluminum’s specific heat and temperature from arbitrary to absolute units. On the answer sheet write the value of the coefficient β of conversion from arbitrary units to absolute units:
c_{\mathrm{Al}}(\mathrm{J}/(\mathrm{kg}\ .\mathrm{K})) = β . c_{{Al}} (arb. units).
(2e) Using the results of measurement of the mass of nitrogen evaporated due to cooling of the aluminum cylinder and the normalized graph of the relationship between specific heat and temperature, determine nitrogen’s specific latent heat of evaporation A. Write the value of λ and its uncertainty Δλ on the answer sheet.
(2a) Below is the table showing the the mass of evaporated nitrogen m_{N} versus time:
Table–1–
(2b) Below is the graph of the mass of evaporated nitrogen m_{N}(t) versus time t (all the three stages of the experiment are shown) :
(2c) Applying the ordinary least squares method to Fig. 7 – 14, we can determine nitrogen’s evaporation rates k_{1}and k_{2} , before the immersion of aluminum and after violent boiling, respectively. Nitrogen’s evaporation rate before the immersion of aluminum is
k_{1}=(5.75\pm0.03)\,\mathrm{g/min}.And the evaporation rate after violent boiling ends is
k_{2}=(4.03\pm0.02)\,\mathrm{g/min}.It is obvious from these rates that evaporation rate depends on the amount of nitrogen in the cup. Therefore the evaporation rate during violent boiling due to heat exchange with the environment can be estimated as the average of the evaporation rate before and after violent boiling:
k={\frac{k_{1}+k_{2}}{2}}={\frac{5.75+4.03}{2}}=4.89\,{\mathrm{g/min}}.To determine the mass m^{Al}_{N} consider the time period from t_{1} = 5. 2 min to t_{2}= 7. 8 min.t_{1}is set by the moment of immersion, t_{2}is determined as described in the introduction, or by analyzing the m_{N}(t) dependence. Then
m_{N}^{\mathrm{Al}}=\,(m_{N}(t_{2})-m_{N}(t_{1}))-k(t_{2}-t_{1})
= (95 – 30) – 4.89 ×(7.8 – 5.2)
= 52.3 g.
It should be noted that any other mean, for instance geometric mean, can be used as an estimate of nitrogen’s average evaporation rate due to heat exchange with the environment. The difference between the arithmetic mean and geometric mean will be used as an estimate of the error of average evaporation rate: Δk =± 0.10 g/min.
The uncertainty is
\Delta m_{N}^{\mathrm{Al}}=\Delta m_{N}(t_{2})+\Delta m_{N}(t_{1})+k(t_{2}-t_{1})\left(\frac{\Delta k}{k}+2\,\frac{\Delta t}{t}\right)=2.4\,\mathrm{g}.Then
m_{N}^{Al}=(52.3\pm2.4)\mathrm{g}.(2d) In the 45°C- 65°C temperature range the specific heat of aluminum is approximately constant and equal to the measurement performed in
Part 1:
c_{\mathrm{Al}}=0.90\;\mathrm{kJ}/(\mathrm{kg.K}).In this temperature range, the value of specific heat in arbitrary units is
c_{Al} (arb. units) = 4. 5 arb. units.
Consequently the coefficient of conversion of specific heat from arbitrary units to absolute units, β, is
\beta={\frac{c_{\mathrm{Al}}}{c_{\mathrm{Al}}(\mathrm{arb,~units})}}={\frac{0.90}{4.5}}=0.2 kJ/(kg. K· arb. units).
(2e) The amount of heat transferred from the aluminum cylinder to the liquid nitrogen as the cylinder is cooled down to the temperature T_{N} = 77 K, is equal to
Q=m.\int_{\tau_{\mathrm{N}}}^{\tau_{r}}c_{\mathrm{Al}}(T)\mathrm{d}T.The value of this integral can be found using numerical integration. It can be approximated as the area under the c_{Al}(T) curve. In our experiment, the number of cells under the curve is N = 311 ± 1 and each cell represents 0.5 J/(g· K), then
\int_{\tau_{N}}^{\tau_{r}}c_{\mathrm{Al}}(T)\operatorname{d}\!t=155.5 ~\mathsf{{J}}\!/\operatorname{g}\!.Then the amount of heat released is
Q = 69 × 155. 5 = 10. 73 kJ.
The value of specific latent heat of nitrogen’s evaporation can be found from the heat balance equation,
m_{N}^{Al}·\lambda=Q.Thus we finally get
\lambda={\frac{Q}{m_{\mathrm{N}}^{\mathrm{Al}}}}={\frac{10.3\,\mathrm{kJ}}{52.3\,~\mathrm{g}}}=205\,~\mathrm{J/g}\Delta\lambda=\lambda\Bigl(\frac{\Delta c_{\mathrm{Al}}}{c_{\mathrm{Al}}}+\frac{\Delta N}{N}+\frac{\Delta m_{\mathrm{N}}^{\mathrm{Al}}}{m_{N}^{\mathrm{Al}}}+\frac{\Delta m}{m}\Bigr)=60\;\mathrm{J/g}.
Table–1–
| N | M(g) | t(min. sec) | t(min) | m_{N}({g}) |
| 1 | 250 | 0 | 0 | 0 |
| 2 | 248 | 0.19 | 0.3 | 2 |
| 3 | 246 | 0.39 | 0.6 | 4 |
| 4 | 244 | 0.59 | 1 | 6 |
| 5 | 242 | 1.19 | 1.3 | 8 |
| 6 | 240 | 1.38 | 1.6 | 10 |
| 7 | 238 | 2. 00 | 2 | 12 |
| 8 | 236 | 2.19 | 2.3 | 14 |
| 9 | 234 | 2.41 | 2.7 | 16 |
| 10 | 232 | 3.02 | 3 | 18 |
| 11 | 230 | 3.22 | 3.4 | 20 |
| 12 | 228 | 3.44 | 3.7 | 22 |
| 13 | 226 | 4.06 | 4.1 | 24 |
| 14 | 224 | 4.28 | 4.5 | 26 |
| 15 | 222 | 4. 50 | 4.8 | 28 |
| 16 | 220 | 5.13 | 5.2 | 30 |
| 17 | 274 | 5.52 | 5.9 | 45 |
| 18 | 269 | 6. 00 | 6 | 50 |
| 19 | 264 | 6.07 | 6.1 | 55 |
| 20 | 259 | 6.18 | 6.3 | 60 |
| 21 | 254 | 6. 30 | 6.5 | 65 |
| 22 | 249 | 6.41 | 6.7 | 70 |
| 23 | 244 | 6.54 | 6.9 | 75 |
| 24 | 239 | 7.09 | 7.1 | 80 |
| 25 | 234 | 7.25 | 7.4 | 85 |
| 26 | 229 | 7. 40 | 7.7 | 90 |
| 27 | 224 | 7.48 | 7.8 | 95 |
| 28 | 222 | 8.06 | 8.1 | 97 |
Cont.
| N | M(g) | t(min. sec) | t(min) | m_{N}({g}) |
| 29 | 219 | 8.49 | 8.8 | 100 |
| 30 | 217 | 9.16 | 9.3 | 102 |
| 31 | 215 | 9.44 | 9.7 | 104 |
| 32 | 213 | 10.14 | 10.2 | 106 |
| 33 | 211 | 10.44 | 10.7 | 108 |
| 34 | 209 | 11.14 | 11.2 | 110 |
| 35 | 207 | 11.45 | 11.7 | 112 |
| 36 | 205 | 12.13 | 12.2 | 114 |
| 37 | 203 | 12.43 | 12.7 | 116 |
| 38 | 201 | 13.14 | 13.2 | 118 |
| 39 | 199 | 13.46 | 13.7 | 120 |
| 40 | 197 | 14.16 | 14.3 | 122 |
| 41 | 195 | 14.5 | 14.8 | 124 |