7
Question
You have m kgs of a certain liquid in a vessel, which is heated till there is a temperature change of ΔT. If the coefficient of volume expansion is γ, what is the density of the liquid after heating, if the initial density was ρ?
Ik the answer...but how did...in the last step did
ρ/(1+y∆T) changes to ρ(1-y∆T)
Please explain this step.
Thank you!
You have m kgs of a certain liquid in a vessel, which is heated till there is a temperature change of ΔT. If the coefficient of volume expansion is γ, what is the density of the liquid after heating, if the initial density was ρ?
Ik the answer...but how did...in the last step did
ρ/(1+y∆T) changes to ρ(1-y∆T)
Please explain this step.
Thank you!
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Solution
Here (a+b)^-1 relation is using.
(1+b)^-1=1-bt+(bt^2)-(bt^3)+(bt^4).......etc...
Compare this with the equation you shown...
(1+y∆T)^-1= 1-(y∆T)+[(y∆T)^2 ]+.....etc.
there ∆T will be a small change approximately equals to zero or a decimal like 0.001 or 0.0001...In coming terms ,power of this term will be 2,3,4,.....etc....
we know square(raised to 2/ ^2) of these terms will be more small like square of 10^-3 is 10^-6 ,cube of 10^-3 is 10^-9 which is more smaller than the 10^-3...Similarly in above equation, the square ,cube terms are neglecting due to zero approximate value...
so (1-y∆T) is only taking...
Like if satisfied
(1+b)^-1=1-bt+(bt^2)-(bt^3)+(bt^4).......etc...
Compare this with the equation you shown...
(1+y∆T)^-1= 1-(y∆T)+[(y∆T)^2 ]+.....etc.
there ∆T will be a small change approximately equals to zero or a decimal like 0.001 or 0.0001...In coming terms ,power of this term will be 2,3,4,.....etc....
we know square(raised to 2/ ^2) of these terms will be more small like square of 10^-3 is 10^-6 ,cube of 10^-3 is 10^-9 which is more smaller than the 10^-3...Similarly in above equation, the square ,cube terms are neglecting due to zero approximate value...
so (1-y∆T) is only taking...
Like if satisfied
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