The density of a substance at 100 degrees centigrade is 1.22 g/(cm^3) and at 140 degrees centigrade is 0.95 g/(cm^3). Estimate the density at 115 degrees centigrade using linear interpolation. Observe significant digits.
density@115=density at 100 + (115-100)(.95-1.22)/(140-100)
= 1.22 + 15*(-.27)/40
compute that.
hey @bobpursley thanks, i got 1.11875 for your "1.22 + 15*(-.27)/40"
I didn't worry with sig digits, you need to do that, and it is a bit tricky. For instance, how many sig digits are in the temperatures?
is the answer that i got is correct??
no, no close to six digits. Follow the rules.
6 sig digits
The linear interpolation formula is given as:
y = y1 + ( x − x1 ) ∙ ( y2 − y1 ) / ( x2 − x1 )
in this case:
ρ = ρ1 + ( t − t1 ) ( ρ2 − ρ1 ) / ( t2 − t1 )
so
t1 = 100°C , ρ1 = 1.22 g / cm³ , t2 = 140°C , ρ2 = 0.95 g / cm³
ρ = ρ1 + ( t − t1 ) ( ρ2 − ρ1 ) / ( t2 − t1 )
ρ = 1.22 + ( t − 100 ) ∙ ( 0.95 − 1.22 ) / ( 140 − 100 )
ρ = 1.22 + ( t − 100 ) ∙ ( - 0.27 ) / 40
ρ = 1.22 + ( t − 100 ) ∙ ( - 0,00675 )
ρ = 1.22 + ( - 0,00675 ) ∙ t − 100 ∙ ( - 0.00675 )
ρ = 1.22 - 0.00675 t + 0.675
ρ = 1.895 - 0.00675 t
ρ = - 0.00675 t + 1.895
The density at 115°C
ρ = - 0.00675 ∙ 115 + 1.895 = - 0.77625 + 1.895 = 1.11875 g / cm³
ρ = 1.12 g / cm³
The result is rounded on two decimal places.
Interpolation can not give more precision than original values.
The original values are given on two decimal places.
So the result needs to be rounded on two decimal places.