Chapter 13, Problem 11P

To determine

To Find: The density of water.

Answer to Problem 11P

The density of water at $94°\text{C}$ is $982.32{\text{ kg / m}}^{\text{3}}$ .

Explanation of Solution

Given info:

Initial temperature $t_i=4°\text{C}$ .

Final temperature $t_f=94°\text{C}$ .

Density of water at $4°\text{C,}{\rho }_w=1\times {10}^3{\text{ kg / m}}^{\text{3}}$ .

Volumetric temperature expansion coefficient for water is ${\beta }_{water}=0.0002\frac{{\text{m}}^{\text{3}}}{{\text{m}}^{\text{3}}\times °\text{C}}$ .

Formula used:

New density at higher temperature :

  ${\rho }_{new}=\frac{{\rho }_w}{1+\beta \times (t_f-t_i)}$

Calculation:

Substitute the given values the density at $94\text{ °C}$ is,

  $\begin{array}{c}{\rho }_{new}=\frac{1\times {10}^3}{1+\left(0.0002\times (94-90)\right)}\\ =\frac{1000}{1.018}\\ =982.32{\text{ kg / m}}^{\text{3}}\end{array}$

Conclusion:

Thus, the density of water at $94°\text{C}$ is $982.32{\text{ kg / m}}^{\text{3}}$