S3 phy Q!! 急!

Assume that the thermometer reading grows linearly with temperature.


圖片參考：http://imgcld.yimg.com/8/n/HA00430218/o/20140107222914.jpg


2014-01-07 23:24:31 補充：
郭老師，我知道這是中學的物理題，你我小時候都做過。
所以知道解法是這樣。

但其實我個人認為是不對的。

因為由 0°C 到 100°C 的增幅不是一個比例上升，你我皆知有 absolute zero 的存在。在於數據分類的角度看，degree Celsius 更經常被用作為 interval type data 的常用例子，即非 ratio type data。（可見 level of measurement 的資料）

所以我作答時故意特地說明清楚 assume increase linearly.

2014-01-07 23:28:09 補充：
我的意思是，由 0°C升到20°C，並不是「由 0°C升到10°C」的兩倍，因為我覺得溫度計中的液體澎脹是根據 -273°C = 0K 的 Kelvin scale。

所以個人覺得這些中學題是不對的，不知道有否理解錯誤～

望 郭老師 和 天同知識長 可以講解一下～

2014-01-07 23:34:42 補充：
Question 2

The heating coil makes 2 kg of water increase 40-15 = 25°C in a minute.
That means, in one minute, it releases 2*4200*25 = 210000 J of energy.

The Power is Energy/Time = 210000/60 = 3500 W = 3.5 kW

The answer is B.

2014-01-08 15:41:26 補充：
Question 3

Let C be the specific heat capacity (SHC) of copper.
Let x and y be the temperature rise of X and that of Y.
The amount of energy is the same.

Energy = SHC * mass * temperature change

C * 3 * x = C * 5 * y

x : y = 5 : 3

1. This is just a simple proportion.
True temperature for reading on thermometer to be 40'C = [(40-10)/(90-10)} x 100'C
= 37.5'C

2. Mass of water heated in 1 minute = 2 kg
Hence, mass of water heated up in 1 s = 2/60 kg
Heat gained by water = (2/60) x 4200 x (40 -15) J = 3500 J
Thus, power of heater = 3500 w = 3.5 kw

When the thermometer is placed in steam. It should read 100 but now it is 90.

When the reading is x, the true value should be (x-10)(100/80)

True reading for 40⁰C should be (40)(100/80) = 37.5⁰C

Ans : C

2014-01-07 22:47:34 補充：
批卷貓 的做法很好，如有疑問，把 ? 換為 x 去想好了。

以下提供一個檢查的方法，當温度讀數為 50 度時，剛好為中點，90-50=40。50-10=40。

批卷貓 的方法 : (?-0)/(100-0) = (50-10)/90-10) = 40/80

? = (40/80)(100) = 50

(x-10)(100/80) = (50-10)(100/80) = 50