Question: (a) Assuming that water has a density of exactly 1 gm/cm³, find the mass of one cubic meter of water in kilograms. (b) Suppose that it takes 10.0 h to drain a container of 5700 m³of water. What is the “mass flow rate,” in kilograms per second, of water from the container?

The density of water, $\rho =1g/c{m}^3$

The time to drain a container, t = 10,0 h

The volume of water,$v=5700m^3$

The density of a material (in this case water) is given by mass per unit volume. The mass flow rate is defined as the mass of a fluid (water) passing through a point per unit of time.

The expression for density is given as:

$\rho =\frac{m}{v}$ … (i)

Here, $\rho$is the density, m is the mass and v is the volume.

The expression for the mass flow rate is given as:

$m=\frac{m}{t}$ … (ii)

Here, m is the mass flow rate and t is the time

First, you have to convert the density of water from$gm/c{m}^{3}intokg/m^3$

$1\frac{gm}{c{m}^3}=1000\frac{kg}{m^3}$

Using equation (i), the mass of the water is calculated as:

$$\begin{gathered} m=\rho \times v \\ =1000kg/m^3\times 1m^3 \\ =1000kg \end{gathered}$$

Thus, the mass of one cubic meter of water is 1000 kg .

Using equation (i), the total mass of water to be drained is calculated as:

$$\begin{gathered} M=\rho \times V \\ =1000\frac{Kg}{m^3}\times 5700m^3 \\ =5700\times {10}^{3}kg \end{gathered}$$

Now, to find the mass flow rate in kilogram per second, convert time (10.0 h) to seconds.

$$\begin{gathered} t=10h\times \frac{3600s}{1h} \\ =36000s \end{gathered}$$

Using equation (ii), the mass flow rate is calculated as:

$$\begin{gathered} m=\frac{m}{t} \\ =\frac{5700\times {10}^{3}kg}{36000s} \\ =158.33kg/s \\ \approx 158kg/s \end{gathered}$$

Thus, the mass flow rate is 158 kg/s .