Question

Volume of Water Between 0°C and 30°C, the volume V (in cubic centimeters) of 1 kg of water at a temperature T is given by the formula

$V=999.87-0.06426T+.0085043T^2-.0000679T^3$

Find the temperature at which the volume of 1 kg of water is a minimum

Short answer

The temperature at which volume is minimum is$79.48\mathrm{K}$

Step-by-step solution

Step 1. Derivative concept to obtain maxima or minima.

To obtain the values minimum or maximum values the first derivative is zero and second derivative is positive and negative respectively

Step 2. checks for first derivative.

Given

$V=999.87-.06426T+.0085043T^2-.0000679T^3$

$\frac{dV}{dT}=0$

$\frac{dV}{dT}=0.017T-.06426-.002037T^2$

Step 3. rearrangement of equation and equating with zero.

$\begin{array}{l}{T}^2-83.45T+315.46=0\\ T=\frac{83.45\pm 75.51}{2}\\ T=79.48,3.97\end{array}$

Step 4. check for second order derivative.

$\frac{d^{2}V}{d{T}^2}=2T-83.45$

Step 5. substuting the values to be positive and negative.

If positive case of minimum and if negative case of maximum

Step 6. values check.

$\begin{array}{l}T=2(79.48)-83.45\\ T=75.51\end{array}$

Positive

Therefore values are minimum at$T=75.51$