Answer:
T = 3.967 C
Step-by-step explanation:
Density = mass / volume
Use the mass = 1kg and volume as the equation given V, we will come up with the following equation
D = 1 / 999.87−0.06426T+0.0085043T^2−0.0000679T^3
= (999.87−0.06426T+0.0085043T^2−0.0000679T^3)^-1
Find the first derivative of D with respect to temperature T
dD/dT = ![\dfrac{70000000\left(291T^2-24298T+91800\right)}{\left(679T^3-85043T^2+642600T-9998700000\right)^2}](https://tex.z-dn.net/?f=%5Cdfrac%7B70000000%5Cleft%28291T%5E2-24298T%2B91800%5Cright%29%7D%7B%5Cleft%28679T%5E3-85043T%5E2%2B642600T-9998700000%5Cright%29%5E2%7D)
Let dD/dT = 0 to find the critical value we will get
= 0
Using formula of quadratic, we get the roots:
T = 79.53 and T = 3.967
Since the temperature is only between 0 and 30, pick T = 3.967
Find 2nd derivative to check whether the equation will have maximum value:
![-\dfrac{140000000\left(395178T^4-65993368T^3+3286558821T^2+2886200857800T-121415215620000\right)}{\left(679T^3-85043T^2+642600T-9998700000\right)^3}](https://tex.z-dn.net/?f=-%5Cdfrac%7B140000000%5Cleft%28395178T%5E4-65993368T%5E3%2B3286558821T%5E2%2B2886200857800T-121415215620000%5Cright%29%7D%7B%5Cleft%28679T%5E3-85043T%5E2%2B642600T-9998700000%5Cright%29%5E3%7D)
Substituting the value with T=3.967,
d2D/dT2 = -1.54 x 10^(-8) a negative value. Hence It is a maximum value
Substitute T =3.967 into equation V, we get V = 0.001 i.e. the volume when the the density is the highest is at 0.001 m3 with density of
D = 1/0.001 = 1000 kg/m3
Therefore T = 3.967 C