1973-01-24 — Page 22

1913英文中學會考試題預習專欄

of finding the specific heat of a substance is called the method of

頁二第張六第二日一廿月二十年子壬夏

WAH KIU YAT PO

郭日僑華

三期星

日四廿月一年三七九一屦公年二十六國民華中育教僑華

we find 1.5 x 15.5

1978 -

288

1.5 x 15.5 x 278

288 x 14.2

em. of

中文中學會考試題預習專欄 中國歷史科(十四) ·贯一民·

灣道英文書院主編

物理科(十四)

Physics (14)

Calorimetry.

Calorimetry is the measurement of heat. We have seen that heat i8 one form of energy due to the thermal motion of the particles in the substance... It is also measured in joules the same as other kinds of energy

Two very important units or heat came into general use were the calorie (cal) and the kilocalor -le(kcal).

The calorie is defined as the quantity of heat necessary to rais the temperature of 1 gm. of water by 1°C. The kilocalorie is that quantity of heat required to raise på 1 kg of water by 100.

kcal - 1000 cal.

Before proceeding to the measure- nents, of heat, we define same basic quantities of a substance.. (1) Heat capacity.

The heat capacity of a body is the heat required to raise its kamnarature hv. 100.

The thermal unit of heat capaci -ty is in cal/°C. In SI unit, it is in J/°C. A body has a large heat capacity means that a large quantity of heat must be supplied in order to raise the temperature by 190.

(2) Specific heat capacity or specific heat. The specific heat capacity of a body is defined as the quantity of heat required to raise 1 kg mass of the body by 1°0

The specific heat capacity of water is 1 cal/gm °C in themal unit and is 4200 J/kg oC in SI 11 unit. Thus 4200 J of heat energy must be supplied to 1 kg of water in order to raise its temperature though 100 From the specific hea capacity of water in thermal unit and SI unit, we can see that

cal. 4.2 3.

Suppose ʼn kg of a body at temper- ature 0,°0 is raised to a temper- ature 0,00. If C is the specific heat capacity of the body, then the total amount of heat absorbed ay the body is given by

Joules

This is the general formula for Computing the heat absorbed or given out by the body.

Thus the heat required to boil 10 kg of water at 000 i

Q = 10 x 4200 x 100 = 4.2 x 10°J. because the boiling point of wate: is 10000. When two substances (on g a solid and the other a being a liquid or both being liquids) at different temperatures are mixed, heat will be given out by the hot body and absorbed by the cold bod until both substances reach a common temperature. Some heat may be gained from, or lost to the surroundings. But by taking suit- able precautions, the amount of heat gained from the surrounding or heat loss can be reduced to a very extent. Thus to a good appro ximation, we can assume that there is no heat loss or gain of the system of interest. Then by conser vation of energy, the heat lost-

by the hot body must be totally absorbed by the cold body at the final steady state of the system.

If M1 kg of a body at 6j°C is mixed with M2 kg of a liquid at

2°C (where 0? 2) and the final common temperature reached is 900 then the heat lost by the hot bod

190, = M101(01 *)

and the heat gained by the cold. body is

= MC (0 06)

Where C1 and U2 are the specific heat of the bodies respectively. Assuming no heat loss, we have My 01 (☺q = 6) = M2ɑ2 (0 - @2). From this, if C, is known, and M1, M2, ẹ, e2 and ✪ are measured C2 can be determined. This method

mixture. The detail of the experi ment will be discussed later, Examples.

(1) 200 gm. of boiling water are: quickly poured into a thick copper vessel of mass 2000 gm. and o origi- hally at 1800. After stirring, the water is observed to remain steady At 60°0. Calculate

(1)

The thermal capacity of the. resset.”

(ii) The specific heat of copper. The heat lost by the boiling water

200 T000

S

x 4200(100–60)

840 x 40

33600 J.

Assume all these heat energy is used to raise the temperature of the vessel, then the heat absorbed. by the vessel is

22

where copper then

20(60–18)

840 J

is the specific heat of

33600 400 J/kg 0.

840

The thermal capacity of the vessel is -MC-400 x 2 =800 J/°C.

(2) A bath containing 100 kg or water at 60°C. Hot and cold taps are then turned on to deliver

- kg/sec. each at temperatures of

70°C and 1000 respectively. Find the time taken for the water in the bath to become 45°C. Assume complete mixing of water and ignore heat losses. The rate of delivery of water

1

kg/sec. 20 kg/min.

Let t be the time taken in minutes for the water in the bath to reach 4500.

In this time t, the amount of water delivered by the taps is 20t kg each. The heat loss hot water from the tape is

20t x 4200(70-45)

20 x 4200 x 25% J.

the

The heat loss by the wat bath is

the

= 100 x 4200(60-45) = 100 x 4200 x 15 J

The heat absorbed by the cold vater from the tape is o

20tx4200(45-10). 20 x 4200 x 35t.

no heat is lost,

L.e. 20x4200x25t+100x4200x1

120 x 4200 x 35t

or 20 x 4200 x 10t – 100 x 4200x15

t = 7.5 min.

3. (An example in the perfect gas equation.) A barometer tube 90 cm. long contains some air above the mercury. The reading is 74.5 cm. when the true barometric height is 76 gm and the temperature is 1500 If the temperature is 5oC and the reading observed is 75.8 cm., What is the true pressure?

Fig (1) represents the true barometric height while (ii) represents the faulty barometer. Let A be the cross-section of the

tube in cm. The air pressure inside the tube is".

P = 76-74.5= 1.5 cm. or mercury

Remember that when the perfect gas equation is applied, the temperat- ure must be in absolute scale. Thus the initial air temperature 18

273 +15 2880K

When the temperatu

273 =278°K

T

The air pressure is

is 500

mercury = 1.58 cm. of mercury.

This is the air pressure above

mercury inside the tube. The true atmospheric pressure is 75.8 +1.58.

77.38 cm. of mercury. Exercise 5.

(1) A mercury barometer contains some air. On a day when the true barometer reading is 75 cm, and the temperature 12o0, the faulty barometer reads 74 cm. What is the true barometer reads 75 cm. and the temperature is 21°C. The

length of the faulty barometer is 1 m. above the mercury surface in the trough.

(2) Two glass pulus ur volumes: 500 cm3 and 100 cm3 are connected

by a narrow glass tube or negligible volume. When the apparatus is sealed off the

pressure of the air inside is 70 om. of mercury and the temperature is 2000. What will the pressure become if the 100 cc. bulb is kept at 2000 and the other is heated to 100°07

(3) Calculate the specific heat of marble from the following data Mass of aluminium vessel-

34.55 gm.

lass of aluminium vessel with cold water

93.84 g.

temperature of aluminium vessel with cold water 1500

temperature of heater containing narble 10000, /

Final corrected temperature of mixture of water and marble in aluminium vessel - 2400

mass of vessel, water and marble. = 134.44 gm.

3pecific heat or aluminium 2840J/kg°C.

(4) A copper vessel of mass 42.88

m.contains water 58.71 gm. at 1000. Some lead pellets from a steam heater is poured into the ressel. What is the final temper iture of the mixture? The mass of ressel, water and lead pellets is 1 252 – 89 gm., and specific heats opper 400, lead 130 J/kg°C.

SOLUTION TO ex.

Let V be the volume of the- balloon, Since Vof the balloon

18 submerged, the weight of water displaced is

V kg.wt.

If the ballooniis immersed atza lepth h below the surface, and is again at equilibrium, the volume of the balloon must becoem.

V cum

The original pressure of air in the balloon is atmospheric which is equivalent to 10 m. of water volumn. When the balloon is at a depth ha, the pressure of air in the balloon is (10+h) w. of water By Boyle's law,

10 x V = — v (70+h)

ans

2. The air pressure in the balloon. when it is at the bottom of the lake ish

76 x 13.649) m. of water

"00:

19.3 m. of water.

The atmospheric pressure is 10.3 m Let V be the volume of the bubble when it is just below the surface. Then by applying Boyle's law, 10.3V 19.3 x 1.5 x 10

V1 = 2.8 x 10 litre

3. Let y be the length of the diving bell and A its cross-sant- ion."

Suppose one bell is lowered into the sea until it is at a depth of x m. from the surface, measured. from the top of the bell. Then initial volume of the bell

=yА cu.III..

and initial air pressure.

76 % 13600

my of sea water

Or sea-water

Find volume of air inside the bell

ind final pressure of air

·預習題答案 T :甲部:選擇題

乙部:問答題

~期

10 C

(一)試說明售代徭望割據形成之因及其對臺堂之影响•

答:(一)浙江之间的形成。

(1) 由於安史之亂平後,各商翩擁兵自宜———唐初只有節座健 之歌,及至安史之氪平後,各鐶之節麗使,權力更大,您 新數州之軍政、財政之大權,日久,便擁兵稱絕,各自蟛 政,儼然一獨立王國。

(2)日於各飙飾嵌使多為融降——當安史之亂時,政府筒求 4971 BEZRAK, STEATROSKESS 安使。如侯希逸爲溜满藏密度使。曹無為池西鐵定健。 21UPINISTER · BUS FES DER BUNAAN DEN BRE DE ERM. 爲城博讯密使。其中尤以盧籍、成機、魏博〨購爲害蔌 烈,彼此互相聾換,不供狀況,目無中央,朝廷亦無可 何

浙爲蘭之影响,

藩鎮為素,使朝廷難於對付,故不得不帳財以兵。

2. 在浙江政權下之地方經濟破證,文化低落,這致日後外族

乘之入一。

3.不僅影响唐之簇亡,而且促成五代十國之局。

)解下列二則及其對審案之影

1:43XP.

2.霞巢之亂,

1.牛平常争:唐朝中葉以後,有所謂「牛李黨爭』,牛黨以牛很開 質,不就可以李德裕為甘。專樣起於意宗元和三年(公元808 年)之科舉考試。當時李吉甫(強裕紀)爲,半價、李宗閱 對策,力話時政之失,於是牛,李從此籍季,彼此各種煮派,瓦 相排斥,爁宗、穆宗、武宗、宣宗四朝,龍嫌四十年之久·道 永黨爭,對唐宝影响颇大,整彼此意氣用事,無神國計民生,反 而宦官藉此㟖懼,且由於彼此眨騎排斥,國家棘核之人材,也爲 之貶議殆癮。對唐審之新亡有很大之關保

·黃巢之亂:黃巢,曹州人民,耒世業變,無因應然不信,還抽 命,乘唐末朝政日腐,天災流行之際,於僖宗載符初,證濮州 王仙芝起蛎。仙芝戰死,單統其余,聲勢浩大。今河南、江西、 浙江、瀾堂、蜜朮、湖南及湖北等省,均受其弱,並攻陷畏安, 與大熷帝。後因單將朱温降,及沙陀人李克用領兵相接,微事始 平。黃巢之亂,前後十年,直接影响唐之覆亡。因朱溫降廣之後 ,助平亂有功,傳宗乃賜其名全忠,自是全忠搬業,結果,新宝 為求所基

式中央及地方官制之組織如傢?與漢代有何不同,其分別黜明之

中央官制:唐之中央官制,以三省為主,三省就是:倪書 強:中書省及門下者。其組織如下

1.尙省:尙書省為全國形高行政機關,不執行政今,其

首長為尚書令,上設吏、戶、體

掌全國政務。............

2:中油省:首長為中書令,不頒發命令。帶及國家大事, ∴由中震雀擬訂、帝豪署,經贏帝同意後,門下

3.門下省百長儒門下街中,罩封象之權。凡中雲省展訂 之政务,由門下省響核,經同意簽署之後,交流賽省教 行。若中赛省之政令,未經門下省簽響,則爲不合法 。又若问下管不同意之政令,可原封最簿。那時中書省 須重新實。

△地方官制唐之地方官制,初佳州、縣二敬,但坚太宗之 時,因山川形勢之便,分全國龠十道(玄宗時艾增速 :),於是成龠酒、州、縣三极。後來又設立府,以;屬州

騷之軍事。將以下則無鄉、陳、、

三)典漢代之不同:漢之中央官制,是三公,即丞相、太尉改

御史大夫。而唐期為三省,即尙書省、中書省及門下省 ̇彼此組織及職實均不同。因爲漢之政務備由丞相府一種關

零管,但唐代則分由三省共同來負責之欲出。

至於地方官制,漢初是郡田制,是採取封建和鄒廠述: 行的制度,但唐代的道、州、縣則是單軌的,不是雙軌並 行的。漢之地方官制,到了東漢時,則演變埝州、祁、解

,又作别除了

何謂租、蒲、調制?有何優點?試确远之。

(眼)糕、開、研制:法初的土地賦稅制度,稱「粗、唐、鋼制

(或稱田配制》。是由北魏的均田制演變而來。我略冰其內容

T

1第一即由租,政府于民以田,民則當業納税于政府。其法

“凡丁男(千八歲》授田百畝,其中八十畝口分田(豐田), 不得實者,百年歸絡後須交還政府。其餘二十款府露永熊田 無須交還政府。民接受田地後,帳丁浩嵗納某二石。

(未完轉入第六張第三頁)

10.1

water

of sea

Again applying Boyle's law, VAX10.1 (10.1+x+2y) XZYA

40.4 - 10.1

Hence

Thus the bell is lowered until the water in it is 3.37 m. below the surface.

کی

Fy for (Qu. 2.)

Since the introduction of air causes a fall of 16 cm. of mercury the initial pressure of air is equal to 16 cm. of mercury, The atmospheric pressure of air is 75 om. as indicated in fig (1). If V be the volume of air intro- duced at atmospheric pressure, then by Boyle few; 25 x 1 x 16 = 75V. or V = 5.33 0.c.