Question

Between 0°C and 30°C, the volume V ( in cubic centimeters) of 1 kg of water at a temperature T is given approximately by the formula V = 999.87 − 0.06426T + 0.0085043T2 − 0.0000679T3 Find the temperature at which water has its maximum density. (Round your answer to four decimal places.)

Answer 1

Answer:

The answer to your question is: T = 4°C

Explanation:

Data

Temperature 0°C to 30°C

mass = 1 kg

V = 999.87 − 0.06426T + 0.0085043T² − 0.0000679T³

T = ? of maximum density

Density = mass / volume, If we look for the maximum density, the volume must be a minimum.

Then, we need to calculate the first and second derivative to find the minimum.

dV/dT = − 0.06426 + 0.017T − 0.000204T²

 Solve the quadratic equation, solutions

 T1 = 3.97       T2 = 79.36  

Second derivative

dV/dT = 0.017 - 0.000408T

Evaluate T1 and T2 in the second derivative to find their sign, if it is positive, there is a minimum and if the sign is negative, it is a maximum.

dV/dT = 0.017 - 0.000408(3.97) =  0.015 positive

dV/dT = 0.017 - 0.000408(79.36) = -0.015 negative

Then

T1 = 3.97 ≈ 4°C is a minimum

T2 = 79.364 is a maximum

Finally, in T1 there is a minimum, and the density will be the highest, becuase the volume will be minimum.

Answer 2

Answer:

temperature is  3.9665

Explanation:

volume is given in the form of equation with respect to temperature

$V = 999.87 − 0.06426T + 0.0085043T^2 − 0.0000679T^3$

we know that, Density is max when volume is minimum.

therefore find the derivatives of the above equation:

V' = -0.0002037 T^2 + 0.0170086 T - 0.06426

for critical point set above {dv/dt = 0} equation equal to zero

-0.0002037 T^2 + 0.0170086 T - 0.06426 = 0

by using below formula find the root of above equation

$T = \frac{ -b \pm \sqrt{ b^2 - 4ac}}{2(a)}$

$T = \frac { - 0.0170086 \pm \sqrt { 0.0170086 ^ { 2 } - 4 ( - 0.0002037 ) ( - 0.06426 ) } } { 2 ( - 0.0002037 ) }$

T≈3.9665, 79.5318

Don't consider 79.5318, as this value lie  outside the given range i.e. 0°C and 30°C.

Hence, temperature is  3.9665