Question

Problem: A reservoir of carbon tetrachloride (CCl4) has a mass of 500 kg and a volume of 0.315m3 . Find the carbon tetrachloride weight, density, specific weight, specific volume and specific gravity. It is given that the specific weight of water at 4 0C is 9810N/m3.

Answer 1

Answer on Question 52989, Physics, Mechanics | Kinematics | Dynamics

Question:

A reservoir of carbon tetrachloride $(CCl_4)$ has a mass of $500kg$ and a volume of $0.315m^3$. Find the carbon tetrachloride weight, density, specific weight, specific volume and specific gravity. It is given that the specific weight of water at $4{}^{\circ}\mathrm{C}$ is $9810N/m^3$.

Solution:

1) Let's find the weight of carbon tetrachloride:

$$W = mg = 500kg \cdot 9.81\,m/s^2 = 4905N = 4.905kN.$$

2) The density of carbon tetrachloride is its mass per unit volume:

$$\rho = m/V = 500\,kg/0.315m^3 = 1587\,kg/m^3.$$

3) The specific weight of carbon tetrachloride is its weight per unit volume:

$$\gamma = W/V = 4.905 \cdot 10^3N/0.315m^3 = 15.57\,kN/m^3.$$

4) The specific volume of carbon tetrachloride is the ratio of its volume to its mass:

$$v = V/m = \rho^{-1} = (1587\,kg/m^3)^{-1} = 0.00063\,m^3/kg.$$

5) The specific gravity is defined as the ratio of the density of fluid to the density of a standard fluid. For liquids, the standard fluid is taken water. So, let's obtain the specific gravity of carbon tetrachloride:

$$s.g. = \rho / \rho_{water\,at\,4{}^{\circ}\mathrm{C}} = 1587\,kg/m^3 / 1000\,kg/m^3 = 1.59.$$

But, we can also define the specific gravity as the ratio of the specific weight of fluid to the specific weight of a standard fluid:

$$s.g. = \gamma / \gamma_{water\,at\,4{}^{\circ}\mathrm{C}} = 15.57 \cdot 10^3N/m^3 / 9.81 \cdot 10^3N/m^3 = 1.59.$$

Answer:

1) $W = 4.905kN$.

2) $\rho = 1587\,kg/m^3$.

3) $\gamma = 15.57\,kN/m^3$.

4) $v = 0.00063\,m^3/kg$.

5) $s.g. = 1.59$.

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