the thread of mercury in a thermometer is 17cm when placed above boiling water, and 9cm when dipped in melting ice. how long would the thread of mercury become if placed in a liquid with a temperature of 75 degree celsius.
is there a method or formula for this?
Yes there is a method. Assume a linear relationship. A temperature range of 100 C (from melting ice to boiling water) moves the mercury column 17-9 = 8 mm
A 75 degree temperature is 3/4 of the way from the 0 degree mark to the 100 degree mark. That is 6 cm past the 0 C mark, or 15 cm.
If you want to use an equation, let X = distance from the bottom of the "thread" of mercury
X = a T + b is the linear relationship.
a and b are constants you must determine.
When T = 0, X = 9, so b = 9
When T = 100, X = 17, so
17 = a*100 + 9
8 = 100 a
a = 0.08 cm per degree.
X = 9 + 0.08 T
When T = 75 C, X = 9 + (75*0.8) = 15 cm
okay... but i think i know another formular which is basically the same, but thanks anyway!
Ggghhhfg
For i think the answer is 12.75cm if we go by ratio and proportion;
100℃=17cm
75℃= x
Cross multiply 75x17 =1275℃/cm
100℃
Degrees celcius will go remain with cm,then 100 into 1275 is
12.75Cm