find the sum of the mass of this container ($d_{water} = 1 g/cm^3$)

Question

The mass of container is $50$g when it is empty. When we complete this container with water, its mass is becoming $130$g. If we unload half of this water and put a liquid which its density is $2 g/cm^3$ instead of this water, find the sum of the mass of this container ($d_{water} = 1 g/cm^3$)

I currently don't have any idea about this question. Can you assist?

Answer 1

Take $m_{c},V_{c},d_{w},m_{total}$ as mass and volume of the container, density of water and mass of the filled up container respectively. We know that $m_{c}+V_{c}d_{w}=m_{total}$ where $m_{c}=50g$, $m_{total}=130g$ and $d_{w}=1g/cm3$. Here we calculate $V_{c}=80cm3$. The new mass after filling the container by half water and half the other liquid which has two times more density than if water, we calculate $m_{new}=m_{c}+ {V_{c}\over 2}d_{w}+{V_{c}\over 2}2d_{w}=m_{total}+{V_{c}\over 2}d_{w}=130g+40*1g=170g$

Answer 2

Hint:

So, the container requires $80$g water to be full, hence its volume is $80$ cm$^3$

Now, $40$ cm$^3$ of water (i.e., $40$g of water) is replaced by $40$ cm$^3$ of another liquid whose mass will be $40\cdot2$g$=80$g .

Answer 3

This container can contain 80 grams of water. The density of water is $1g/cm^3$. So, this container has a capacity of $80cm^3$. The mass of the container is:

$$ m_1 = 50g $$

The half of the container is filled with water. The weight of the water is:

$$ m_2 = 80cm^3 / 2 \times 1g/cm^3 = 40g $$

The other half of the container is filled with a liquid with density $2g/cm^3$. The weight of this liquid is:

$$ m_3 = 40cm^3 \times 2g/cm^3 = 80g $$

The total mass of the filled container is:

$$ m = m_1 + m_2 + m_3 = 50g + 40g + 80g = 170g $$