Between 0 degrees Celsius and 30 degrees Celsius, the volume V (in cubic centimeters) of 1 kg of water at a temperature T is giv

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Between 0 degrees Celsius and 30 degrees Celsius, the volume V (in cubic centimeters) of 1 kg of water at a temperature T is giv

en approximately by the formula V=999.87−0.06426T+0.0085043T2−0.0000679T3 Find the temperature at which water has its maximum density. Temperature = degrees Celsius

Mathematics

1 answer:

PIT_PIT [12.4K] · Answer 1 · 2026-03-05T22:50:55+03:00

Answer:

T = 3.967 °C

Step-by-step explanation:

The formula for density is mass divided by volume.

Using a mass of 1kg and applying the given volume formula V, we arrive at:

D = 1 / (999.87 - 0.06426T + 0.0085043T² - 0.0000679T³)

= (999.87 - 0.06426T + 0.0085043T² - 0.0000679T³)^-1

To find the first derivative of D in relation to temperature T:

dD/dT = $\dfrac{70000000\left(291T^2-24298T+91800\right)}{\left(679T^3-85043T^2+642600T-9998700000\right)^2}$

Setting dD/dT = 0 to determine critical points results in:

$\dfrac{70000000\left(291T^2-24298T+91800\right)}$ = 0

By applying the quadratic formula, we uncover the roots:

T = 79.53 and T = 3.967

Given that temperature is constrained between 0 and 30, we select T = 3.967.

Next, we calculate the second derivative to ascertain if it will yield a maximum:

$-\dfrac{140000000\left(395178T^4-65993368T^3+3286558821T^2+2886200857800T-121415215620000\right)}{\left(679T^3-85043T^2+642600T-9998700000\right)^3}$

Substituting T=3.967 into the equation:

d²D/dT² = -1.54 x 10^-8 a negative outcome indicates a maximum.

When we input T = 3.967 into the volume equation, we find V = 0.001, which indicates that the density is highest at a volume of 0.001 m³, yielding a density of

D = 1/0.001 = 1000 kg/m³.

Thus, T = 3.967 °C.