w腊月二十年子壬磨复
1973英文中學會考試題預習本欄
物理科
(十五)
Physics (15)]
Changes of state.
Juos exist in the
solid
liquid or gaseous state. In the solid state, a body has a regular, seometrical structure. Its atoms. or molecules are arranged in a- regular pattem (called the space Lattice), in the liquid state, a nguy has no definite shape but a fixed volume. It adopts itself to the shape of the container but does not expand to fill. The Laquia molecules have random movement just as gaseous molecules But the liquid molecules are still
lose enough to attract one an- other.
The atoms or molecules of a solid vibrate about their mean position in the space lattice. The kinetic energy of the atoms increases with the temperature of the solid. If the solid 19 heated 9 temperature will be reached at. which the attractive force betwee the atoms can no longer hold them in the space lattice. The solid The is then said to be melted. temperature is called the melting point(m.p.) of the solid. The work (the neat energy) necessary Lo break up the space-lattice,
e to melt the solid, is called the latent heat of fusion. In changing from a solid state to a Liquid state, the temperature of the substance remains constant. The Latent heat of fusion can be rephrased as that quantity of hent required to change 1 kg of the solid subetance from the solid state to the liquid state, Lhe temperature remaining con- stant
Similarly the latent heat of “vaporation is the heat required to change 1 kg of the liquid Substance From
from the liquid state to the gaseous state, the temper ature redtaining constant.
Example
The latent heat or ice is 36000 1/kg. This means that 336000 J of energy is required to change 1 kg. of ice 91 0°C to 1 kg of water. 9 the saune temperature.
Latent heat calculation. Lat Ly be sne Latent heat of.
fusion of a certain sólad atance of mass œ kg. From the
definition of latent heat,
1 1/kg.
Thus the total neat: required to change the m-kg solid substance from the solid state to
liquid state 18:
mL, J
Similarly, the heat required Lo
hance the a-kg liquid substance From the liquid s'ate to the gaandus stete 13 2,- ml, wher L
is the latent heat of eva po'ra ti on for vapoursation) of the liquid
su je tance.
Example 2.
Calculate the total heat requi red to change 10 kg of ice at -10°C to Bleam at 400°C
(latent neat of fusion of ice is 336000 J/kg 4 latent heat of
evaporation of 001ting water 19 2,260,000 U/kg.)
The problem involves the following Stages:
1000 Ice
water
1000C water
Steam
0°C 100000
It is also given that the S.H. of
Ice 19 2100 J/kg°C.
ther
mst 10 x 2100 x 10
ale
mst:
210 kJ
10 x 336 kJ = 3360kJ
10 x 4200 x 100
4200 kJ.
10 x 2260 Ku 22600 kJ
The tot. heat required for the change is
* 22600 + 4200 + 3360 + 210
- 30,370 kJ.
Remark:
The same amount of heat is required for the reverse change
n the above example.
WAH KIU YAT PO
報日僑莘
三期星
日一卅月一年三七九一座公年二十六國民中
育教僑華 M
菠希鳳
1 e. 30,370 kJ is required to change 10 kg of steam to ice at -1000. Because the same amount of heat is required to solidify 0°C water as that required to melt 0°C ice. The same is true for vapouri zation and condensation. Example 3
A copper calorimeter weighing 0.1 kg containe 0.15 kg of water at 30°C. Ica pellets already dried are dropper in and the final temperature after stirring is 50. The weight of the calorimeter and its contents, is then 0.30 kg. How much water was put in with the piece of ice?
(Latent heat of ice = 336 kJ/90 and S.H. of copper is 400 J/kgoC. ) Let M be the mass of water on ice. The mass of wet ice = 300 250
= 50 gm
the mass of dry ice
= (50 - M) gm.
Heat given out by the copper calor -imeter and water in cooling from 3000 to 5oC is
0.1 x 400 x 25
x25
0,1x400x25+0.15x1200x25.
670x25 J.
4200
Heat absorbed by ice and water on ice in the change from 0°C to 5°C is
(50-M)x10 ̄ ̄3×336000+(50–M)
x10x4200x25+Mx10
3x4200x255
(50-M)336+ (50-M)+M ×10−3
x4200
(50-M)336+5x42x5
Assume no heat loss, then Q
e. 610 x 25 = (50-M) 336+5x42x5
25 x 628 336 3.3 gm.
50-M =
Determination of specific heats
The following are the most common methods for solids and liquids.
Method of mixtures.
An empty copper calorimeter with stirrer is first weighed. It is then half filled with water and reweighed to obtain the weight of water. The initial temperature of the calorimeter and its content is taken by a thermometer. The solid whose S.H. 18, required is heated on a water bath to a known temper ature. It 18 then taken out of the water bath and rapidly transferred into the copper calorimeter and its content are taken. The mass of the solid is obtained before heat- ing. By using the usually formual Q = mst, the S.H. of the solid can be determined.
main source of error an
eriment is the heat loss during the transfer of the hot solid from the water bath to the calorimeter. Another source 18 the extra weight of water carried by the hot solid from the water bath to the calorimeter.
(2) By cooling (mainly for liquide
This method consite of first plotting a cooling curve of the calorimeter which contain known weight of hot water and then plotting the cooling curve for the liquid whose S.H. 13 required. The liquid used must be of the same volume as the water to make the temperature distribution over the Burface of the calorimeter the Bame in each experiment.
A
ypical cooling curve 18 shown in the fig. below. Bu using Newton's law of cooling, we have
(mg st
where m
)(02
and m are the mass of
liquid, water and copper calori meter respectively and e
and
are their S.H.
Then
From this s
Exercise 6
cam be determined
4. 40 gm. of ice at -20°C and 1000 gm. of molten lead at 400°C are
mixed. Assume no heat loss to the
surroundings find the final state of the mixture.
S.H. 1cel
Tead =
130 J/kg°C.
Latent heats of steam - 2260 kJ/kg Lead = 2.70 kJ/kg.
Latent heat of steam = 2260 kJ / kg Melting point of lead = 328°C.
2. 8 gm. of ice at 0°0 are dropped into a mixture of oil and water at 2000 and are just melted in c
cooling the mixture to 0C. If the 5.H. of ice = 2100 J/kg°C and the total weight of liquid at the end of the experiment is 60 gm. Find the mass of the oil and water in the original mixture. (Latent heat of fusion of ice.
-336 kJ/kg)
Steam from boiling water is
passed into a mixture of ice and] water until the weight of the mixture has increased by 2 gm.→ What the mass of ice which wi11 have been melted:
(Latent heat of vaporization of water = 2260 kJ/kg. 1. of ice
336 kJ/kg).
Solution to ex. 5.
In fig. 1. the air pressure
• 75-74=1 cm. of mercury.
In fig
18-75
2/2
the air pressure
cm. of mercury.
Hatmospheric pressure
lawy
1 x 26. = (H-75)25
or H-75-
1.04
H = 76.04 cm. of mercury
PV
mR
Applying the equation where m mass of the gas and R is the gas constant for 1 kgm of the gas. For fi
373
70 x 600
293
500
100 293
Then
70 x 600
Px 100
293
Solving
P - 35.5 cm. of mer
3. Mase of cold water
| - 93.84 - 34.55 - 59.29 gm.
Initial temperature of cold water in aluminum vessel is 1500. Initral temperature of marble = 10000.
Find temperature or mixtures
2400.
Mass of marble
134:44 93.84 40.6 gm. Heat absorbed by aluminum vessel and cold water
= 34.55 x 10. x 840 x 9+ 59.29
4200 x 9 x 10.
2520
Heat given out by the marble
40.6.9 x 85 x 103.
460 S
1913英文中學會考試題預習專欄
了上期決
堅遊英文書院主編
中文科
宿江邊閣
2(乙)(一)就上也管不便一雙誤
***RBYE(-)
(十四)
杜甫
(六)進行而不相悖——這首示,
— QUARTE
(七)而以唱出之一之誤入 .無會投辉臟之氣——無誤
MARBLEZKA (乙)(一)牵莪錦宋一眼婆
21
有称将長短句十二行肚——谒起
(二)下桀如生铁成——湄蛋字
POWERM
1.病色延山碟
釋詞:瞑色,原色世·廷:伸腿
全句期日不昏暗之色逐漸伸到民間之小徑上。
詞:諾劑:江過閣(西閣):臨近出。水門:門長江
·號入三峡之問声)
全句譯江邊開鐵近城峽口之張門也(三峽;班缺 BM) •
3.海滨巖際戳
15-1- Mike
全句推微雲飄過山頭,停於樂上似有住宿之意。 和国月滨电路
ERCE: LENNA
∵全句银月光手於水面、波浪搖助似翻轉一般估o!
5.鄭知追茂辉
45:3:84•
全句解鵬鶴於黃昏時分,各自歸巢,其追逐飛 T
6.财狼得实喧!
#H:5:ICE •
全句謂山中豺狼,就得食後仍喧
7.不跟愛戰代
全句命因燹密戰火亚廷而不能入睡出 •
8.無力正乾坤
釋司:正:整頓。乾坤;硫天下。
∵全句萌自歐無能力整頓國家,使遠攷承不盘。
(乙)間答:
(-) RAT2449.
答:杜甫,字于美。初唐诗人杜籐習子。生於廣容宗殆天元年,
(七一二)(輕陜西杜款(不號社陵布表,少野老)先武德 居渊鳎。會租佬藝公,終河南鞏縣令。途在發麤定居焉。甫少 小多病。七歲能爲筛文,家餐,過冠再走衣黨,遊於吳楚,
趙間,後蹄東都,彃進士不第。居長安•天蚤十袋,進三 大难膑。玄宗态之。命德制集緊院。天蜜十四載十月,授河西 尉,不淨。安史亂興,甫負貤中奔胤行在,拜拾藏。兩式 光俊,南田為華州司功參軍。李乗官西去,人依嚴錢。武藥局 色度參謀。橡胶工部員外郎。做武本,携家雞川報下•大颗心 代宗年號)五年生,竟以凱卒於谳,岳C湖劑)問。旅残岳關 △(慊憾公飫死耒陽指中,不實!五十九俊有杜工部集行业。
(二)世稱杜特篇詩史,其放徊在100
答:甫博極案,所城時Æ洋活裔,涵古今。四審愛時,可可 怨。天香末年,安史之亂,關中淪陷,文物描到人民困苦 負連,流離填尼。南本鮮家博愛襟慎,以辭歌永民生疾苦。能 安杜甫傳費:「前善陳時事,律可惜深•至千管不少莪,也就 詩史。」沈騏曰:「可以認龄,可以貸史,可以爲磁॰其實時 形瘸跳,不塊古詩人之妾。蔡亦能之僅有者選•」故市 有詩史之稱
(三)咸述杜甫時之風格,與在文學史上之地位?
答:杜甫詩大氣磁碟,使遠雄準,沈糖在心,贊恩又問,風格之斷 ,睥睨今古,加以各體氣工,無架不確。立官忠厚~可拿可很 有「時觀之母。激辣時事,直司無果,鷹「詩與上场暴
·元旗云上湖斷,下裁洗...............推頓網上孤書、雜 徐躕之施職,並得古今之體勢,而統交人之所獵專疟•時人以 來:未有如于笑者。」關羽云:「少陳特 亞犂波澜,取材於 六腐,至其自得之妙,則前設所謂集大成者也。」其在中國文 「學史上之地位,於此可見。宜其篇百代路人之宗師瘤的
(四)誠殺出忠各期之代表人物,風格不同之處?
唐代學那盛,人才輩出,其應約可分為四期:初唐上承 六與選風,而古機雅鍵,一洗精羅帶之氣。以陳于是,沈佺 期為代表。2.盛:作家跳起,包號萬您,有設置做,無美不 球。以杜甫、李白篙代表3.中店:以韓慈、白居易二人爲最時
• •. EHUR SEXK• 4.19
之時,共風格以锁槃權巧,帶有濃厚浪漫色彩。李商融,提升 無矯代表
(五)本篇主旨與內容分析;;
AF備生局描寫宿於城端口之江邊(西閣)之上,竟夕不 旅,以眼前景物之所廁所聞,發嶌吟詠,特找愛確熊導之情
B內容分析:
一二的抽到燕作時分,夜色蒼茫,關之樵劑,漸能宴色撞 沒。而叫關因沉臨城口,故變問之形勢仍願約可見。 三句永第一句「山」字來,描寫入夜後不能入寐,所見之 眼就录,喻自己從於邊閣。四句承第二句「水」字面來, “不是寫時間之北,以孤月喻自己身世之孤獳也
(按第三四句,由「山」説到「水」是「靜」擗3
我们承第二句「水」雨來,描寫入夜後不能入案,所聞之 配音。嘛自己身
六句承第一句「口」字面來,亦是描寫人夜所聞。除時蜀中 :兵亂,仍未寜息。
(以上第五六句由「歲」說到「獸」,是「勱」
七、八句總插上六句所見所聞,能助其受懷宗國之情罐,而 又自刻才力不准,不能玉扶破也。
no heat is 10st,
3.46 5 2520
M
S = 728 J/kg°C.) Mass of lead pellets 252.89-42.68 – 58.71 156.30
gm.
Initial temperature of copper, vessel + water is 1000. Initial temperature of lead pellets is 10000,
Let t be the final temperature the mixtures.
Heat absorbed by copper vessel
42.88x400x{ t-10)+58.71x4200 x(t-10)
Heat given out by the lead pellets
156.3 x 130 (100
Assume no heat loss,
1156.3 x 130(100-t)
(42.88 x 400 58.71 x 4200)
t = 10)
p0400(100-t) = 263300(t-10)
100 t = 12.9(t-10),'
13.9-229:
b= 16.500.