f4 l物理revision

2014-08-04 7:28 am
http://postimg.org/image/k2z3l4aub/

好混亂啊!好難。
更新1:

我想了很久都不明白你條式的answer係點變出黎架?麻煩你!

更新2:

Hence, 380(m1) + 388(m2) = 385.8(m1 + m2) i.e. 380(m1/m2) + 388 = 385.8(m1/m2) + 385.8 (m1/m2) = 0.38 即係無端端m1+m2變左m1/m2? 左邊的式係咪除左m2?

更新3:

hi天同知識長: 想問番(2) option 果度,water condensing on the cooler part of the beaker……d水會唔會流番落去個beaker度?我只係想知甘解得唔得。唔該。

更新4:

天同: 我的題目無故被移除了,可以再答我一次嗎?我一會在上載photo!

更新5:

我在說長題目。。傳熱那個。他說我違規了!

回答 (1)

2014-08-04 8:27 am
✔ 最佳答案
4. Let m1 and m2 be the masses of copper and zinc respectively.
Consider one degree C rise in temperature,
heat absorbed by copper = 380(m1)
heat absorbed by zinc = 388(m2)
heat absorbed by brass = 385.8(m1 + m2)

Hence, 380(m1) + 388(m2) = 385.8(m1 + m2)
i.e. 380(m1/m2) + 388 = 385.8(m1/m2) + 385.8
(m1/m2) = 0.38

5. Let Q be the heat supplied and m be the mass of water vapourized.
L be the latent heat of capourization of water.

Hence, Q = mL
i.e. L = Q/m

Statement 1 is correct.
When Q is overestimated, L(experiment) is higher than L(true).

Statement 2 is correct.
The measured value of m is lower than the true value. Hence, L(experiment) > L(true).

Statement 3 is wrong.
The measured value of m is higher than the true value.
Hence, L(experiment) < L(true)
The answer is option C.

2014-08-04 07:40:48 補充:
Your suppl questions:
380(m1) + 388(m2) = 385.8(m1 + m2)
Divide each term by m2
380(m1/m2) + 388(m2/m2) = 385.8(m1/m2) + 385.8(m2/m2)
380(m1/m2) + 388 = 385.8(m1/m2) + 385.8

2014-08-04 07:45:03 補充:
Q:想問番(2) option 果度,water condensing on the cooler part of the beaker……d水會唔會流番落去個beaker度?我只係想知甘解得唔得。
A: Yes. The condensing water will go back to the beaker. Thus the measured mass of water loss due to evapouration is less than it should be. This leads to an overestimation of the latent heat.


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