Question

Water enters a 4.00-m3 tank at a rate of 6.33 kg/s and is withdrawn at a rate of 3.25 kg/s. The tank is initially half full.

Answer 1

Explanation:

The given data is as follows.

            Volume of tank = 4 $m^{3}$

             Density of water = 1000 $kg/m^{3}$

Since, the tank is initially half-filled. Hence, the volume of water in the tank is calculated as follows.

                        $\frac{1}{2} \times 4 = 2 m^{3}$

Also, density of a substance is equal to its mass divided by its volume. Therefore, initially mass of water in the tank is as follows.

                    Mass = $Density \times initial volume$

                              = $1000 \times 2$

                              = 2000 kg

Whereas mass of water in tank when it is full is as follows.

                     Mass = $Density \times final volume$

                               = $1000 \times 4$

                               = 4000 kg

So, net mass of the fluid to be filled is as follows.

                  Net mass to be filled = Final mass - initial mass

                                                      = 4000 kg - 2000 kg

                                                      = 2000 kg

Mass flow rate $(m_{in})$ = 6.33 kg/s

Mass flow rate $(m_{out})$ = 3.25 kg/s

       Time needed to fill tank = $\frac{\text{net mass to be filled}}{\text{net difference of flow rates}}$

                                       = $\frac{2000 kg}{m_{in} - m_{out}}$

                                       = $\frac{2000 kg}{6.33 kg/s - 3.25 kg/s}$

                                       = 649.35 sec

Thus, we can conclude that 649.35 sec is taken by the tank to overflow.