Q23P - Fundamentals Of Physics

Question

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 m³ of water. What is the “mass flow rate,” in kilograms per second, of water from the container?

Step-by-Step Solution

Answer

Answer

 

1. The mass of one cubic meter of water is 1000 kg .
2. The mass flow rate is 158 kg/s .

 

Step 1: Given data

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

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

The volume of water, $v = 5700 m^3$

Step 2: Understanding the density of material and mass flow rate

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

Step 3: (a) Determination of the mass of cubic meter of water

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

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

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 .

Step 4: (b) Determination of the total mass of water

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.

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

 

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

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

 

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