日七十月一十年牛芮腦業
WAH KIU YAT P/
级
日八廿月二十年六六九一限公年五十五國民蓄
育教僑華
組織和工作} 「學生會的任務、
遠東書院
否生績
學生專文
文院
「否認的事實。 校方與開學之聯絡,以做到上下
一生會曾付出一份聚力,予以支撑,是無可 ... 本厚生會之主要使命有出,对內乃是一務之競務。 員工與同學的避力合作,互助互勉;在度食用。由加入本會之日起,即享有選集》 一個油盤中,質皭城議,無溫流動,此中心 德藝,創们,從珠,雅免蓋複利;凡育,都市,亚州的大商量,被月爾東方之珠的前提是否然全?目前一個小学生的理想是隨友 並無任何外發育期,認通院方全透世事,一次會會案,均可申請加入本會,成抵正式「與工作」爲思邈美忠文如下,封什,生畫風後水的阿因何在? 學府中 員有遵守本會會宮,交會費,及接受驗「人間天做」,人口的爱保守铈計盈三百 通過升中入學試而取得中學位一個中
各組幹事,任期第一年,任期,
,促進同學的罐,響,造、雄之全商發發一下::
「,短造和工作」發炎應交斯下
[東塔院學生會代表團,以『學生會的任
言語,及各風校際活動,使彼此間有 外方面,應邀參與洋猪各大鸡院校之正常計本會之工作計與方針,
主席——協助主席獻一切行政事
次食必是不生一
是就倒,此我榮:一
風
管
以您欲 理 糖永洲;
聯育適日
如學心的餘
於各而筒。
兩方致育家說:「高昂珓乃麟造: 將起見,茲將底况列如左,
-EREKERJ (BRAINS NOT B
| EFFICIENT SCHOOL)、铋镪鮮 RICRS AND MORTAR, MAKE AN
于又說:「徒法不能自行,一由此发來,
此
作工
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可的一種於
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鋼琴 學術·酵館 鼷利 交
步例
我會
人的質詢-
行做實
案按生意的...定的意
和花貴香們,麗香施
當相的,五怪生1名的人目他
央:中反大会人業服
万的色在
屉肚,露入
因法接盖任這驗無因的,於一目的盛的升是甘角
的和生物
爭神,但大家都能彼此蚊解,使我能按
們每
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的出倒,不,又得
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是肚古, 是情須們
. (E)
「學生會任務、
H
·我院創校迄今,这十三號,而其間
「組織與工作」
經緯書院
學經
生緯
凡砍得我校畢生师录中,而隨意恪守韓書院朗牛代表團以「基宇會任務,祖 〔特凯)第二盤孙上季件研射青,
文院
餓們做學
「學生的理莅何在?」取者:「其什【拉曼的分別,以其教育來區,私立書上流动。以上的阻檢,為任何角寶君來,
下的以使
平大伟乡案大
私以位
大學校問?縱使有,又有雄雄,属正的拿私立大專院换的學生是在爲教育而米實的業 想?除了邊到「實現自我」和「楯退一己之下,多麼就耳丨 許真的抱構「讓身就書」的目的而碟,了私立大忠仍有他的存在價值,同時表明,其缺糖大會以景觀上的偏見。 然而,有多少大專學生在宫里高的理麦的樂器,其學生是財會的孤兒。巴尔良 一官他們的最高理想外,其間有機入大,但假什麼私立大裏仍能存在,所以新就易維所及的和無林雪院學生會館
又社外!
難制,接存院
的旺游,牛學影
許有質赠一改源字亦的一的人了一實以然?,會
小學一們有天我
5英文中學會考試題預習專欄
物理科
(九)
·PITYSICS (9)
,梁海明 H.M.Leung
Solution.
15.(a) Measurement of the coefficient of ex pansion of a gas at constant pressure.
An apparatus designed by Lay is used.
in this experiment.
The gas is enclosed in the bulb by con- centrated sulphuric acid to ensure that it is dry and the apparatus is heated by Immersion in a water-bath, We en sure the constancy of pressure by add. ing acid throug: the funnel F or by removing acid. through the siphon S,until
A and Bare at the
Lay *s apparatus S of the gas is: read from the graduations on the tube at B, and the temperature. from the thermometer T. The temperature must not be raised too rapidly otherwise Twill indicate a temperature higher than that of the gas. The water-bath: must be stirred continiously. The bunsen must be Lowered or removed before a reading is taken to avoid difficulty in keeping AB Level. Results should be recorded in a table and a graph 19 plot. ted with volume as ordinate and tempere sture sa abseissa. The graph is approm- imately a straight line, From the dew finition, we obtain,
same level. The volume:
coefficient or cubical expansion
increase in volume
original volume x temp. change
Where originaı volume is the volume of gas at oc. It is found that this. coefficient is a constant which has a value 1/273 per deg.C. The result is - known as Charles law which stated that a fixed mass of any gas expands by V273 of its voluma at 0°C for each Centigrade degree rise in temperature provided the pressure is kept constant.
(b) The pressure of air inside the bar-
ometers 760-750
10
8ince the reading is unchanged, they volume of air inside is unchanged. From Pi
Let P 10
2735 273+20
293 K
P2
T2
Constant Volume Air Thermometer;
Tr thermometer
Br glass bulb
M: fixed mark mercury level
R: adjusting mercury level
Ns millimeter scale
Advantages compared with a mercury thermometer.
2. It is very sensitive because the value of the coefficient of increase of pres
ssure ofa gas is much larger than the coeffialent of cubical expansion of a liquid.i.e. larger changes of pressure are indicated for quite small changes of temperature.
2. The range over which it can be used
is Large--almost down to absolute zero (with helium as the gas) and as high as 1500c (with nitrogen as the gas). But this is impossible in the case of mer-, cury thermometer.
(b) When the bulb was immersed in melting
ice(temperature 0°C) the pressure of the air in the bulb - 75 cm.
When the bulb was immersed in Dolling water temperature 100°C, the pressure of the air in the bulb = 205cm.
When the bulb was immersed in a boil. ing liquid the pressure of the air is the bulb = 112.5 cm.
1050
Pressure
Tempare ture
100%
from the: pressure-temperature graph, similar triangle,
AB represents a pressure of (105-75) cm.
that is 30 cm.
PQ represents a pressure of (112,5-75) om,
that is 37.5 cm,
CB represents a temperature of 100oC and CQ represents the temperature has to be found. Hence,
G HONGKO
30
32.5. 100
100 x 37.5
which
10x293
273 10.7
Real atmospheric pressure
750+ 10.7
760-7
am
Answer: The real atmospheric pressure
will be 760.7mm.
(c)Assuming constant temperature,
Let V =0.3
cu.in.
pressure at the dottom or the Lake
atmospheric pressure → pressure due to the depth of water. 30x13.6
£16.10
12
27
240=ft, of water
ft.for water
atmospheric pressure
30x 13.6
12
ift. or water
ft. of water. volume of air bubble on reaching
surfa
(ou,in.
cu. In.
of the air bubble v111
on reaching.
= 125.
30
answers The temperature of the bolling
liquld is 1259°C
Volume of the pump cylinder= 10x15c.c.
150
1st Stroke - The 10000.c.
of air in the flask expand
Stroke 15
Ama wat on
to fill the flask and the pump and since the temperature is con- stant, Boyles' law can be applied.
=77 cm. 1000
Po=Required pressur
= 1000+150c
1150
PV=P_V2
77x1000
1150
= 67
2nd Stroke: At the end of the first stroke the valve V closes and the air in the pump is expelled through a second valve in the piston as the latter moves. forward. For the second stroke, there fore we begin with the flask filled with air at 67 cm pressure. Hence, P=67
Ct
Py=Required pressure
in em. -1000c.c. V2- 1150
67x1000
1150
$ 58.3
ep
Answers The pressure in the flask wil be 67 cm of mercury after the first stroke and 58.3 cm of mercury after the second stroke.
Topics for revision this weeks
Quantity of heat; the them. ), specific heat, Change of state; eva Latent heats fusion
B.Th.D.. alent, and boiling
Questions:
+個人的內地來對此一月的,由於入微
分部生做 學名的的
分門
17.(a)Define Jspecific heat and water equl-
valent.. b)Calculate the ratio of the water equi
valents of two vessels having the same volume of metal, one made of iron and the other of copper.(s,g.af iron=7.9;
copper=8.9, specific heat of: iron =0.12; of copper=0.09)
of o
200.gm of water were placed in a vacuum flask whose water equivalent wa! 10 -gm. The temperature of the water wal then 16.0c. A small quantity of solid which is sclucle in water was added. After vigorous stirring until all the solid had dissolved, the temperature -was 14.5°C.How much heat was absorbed
when the solid dissolved? (Take the specific heat of the solution as 1.0) 18,(a)Define the term *latent heat of steam and latent heat of fusion of ice (b)A piece of red-hot metal, specific neat
0.11 calories per cm per deg.cmass 21. em,,temperature 710°C, is dropped intr 100gm of water, temperature 0 0. Some steam is given off and after stirring the temperature of the water is found to be 10 C. The remaining water and metal together weigh 120gm.
(1) Hou much heat is lost by the metal、 (11)How much heat is gained by the re-
maining water?
V111) What result do these figures give
for
the latent heat of steam?... eHow much steam Hat 100°C must be passed
a mixture of 100 gm of 100 and 200
water contained in a caionimeteran..
wess 200gm and specific heat 0.1, so that the final temperature of the mixture shall be -10°C?
IEND4
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