Chapter 1, Problem 17P

(a) Compute the order of magnitude of the mass of a bathtub half full of water. (b) Compute the order of magnitude of the mass of a bathtub half full of copper coins.

To determine

The order of magnitude of the mass of a bath tub half full of water.

Answer to Problem 17P

The order of magnitude of the mass of a bath tub half full of water is $\sim {10}^2\text{kg}$.

Explanation of Solution

Section 1:

To determine: The volume of the half full water bathtub.

Answer: The volume of the half full water bathtub is $0.195{\text{m}}^{\text{3}}$.

Assume length of a bathtub is $1.3\text{m}$, breadth of a bathtub is $0.5\text{m}$ and the height of a bathtub is $0.3\text{m}$. The density of water is $1000\text{kg}/{\text{m}}^{\text{3}}$.

Formula to calculate the volume of the half full water bathtub is,

$V=\frac{l\times b\times h}{2}$

• $l$ is the length of a bathtub.
• $b$ is the breadth of a bathtub.
• $h$ is the height of a bathtub.
• $V$ is the volume of volume of the half full water bathtub.

Substitute $1.3\text{m}$ for $l$, $0.5\text{m}$ for $b$ and $0.3\text{m}$ for $h$ to find $V$.

$\begin{array}{c}V=1.3\text{m}\times 0.5\text{m}\times 0.3\text{m}\\ =0.195{\text{m}}^{\text{3}}\end{array}$

Section 2:

To determine: The order of magnitude of the mass of a bath tub half full of water.

Answer: The order of magnitude of the mass of a bath tub half full of water is $\sim {10}^2\text{kg}$.

Assume length of a bathtub is $1.3\text{m}$, breadth of a bathtub is $0.5\text{m}$ and the height of a bathtub is $0.3\text{m}$. The density of water is $1000\text{kg}/{\text{m}}^{\text{3}}$.

Formula to calculate the mass of half full water in the bathtub is,

$m=\rho V$

• $m$ is the mass of the half full water in the bathtub.
• $\rho$ is the density of the water.
• $V$ is the volume of a bathtub.

Substitute $1000\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ and $0.195{\text{m}}^{\text{3}}$ for $V$ to find $m$.

$\begin{array}{c}m=\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(0.195{\text{m}}^{\text{3}}\right)\\ =195\text{kg}\end{array}$

Since, the mass of half full of water is $195\text{kg}$ and can be rewrite as $1.95\times {10}^2\text{kg}$. So, the order of magnitude is $\sim {10}^2\text{kg}$.

Conclusion:

Therefore, the order of magnitude of the mass of a bath tub half full of water is $\sim {10}^2\text{kg}$.

To determine

The order of magnitude of the mass of a bath tub half full of copper coins.

Answer to Problem 17P

The order of magnitude of the mass of a bath tub half full of copper coins is $\sim {10}^3\text{kg}$.

Explanation of Solution

Section 1:

To determine: The volume of the half portion of bathtub.

Answer: The volume of the half portion of bathtub is $0.195{\text{m}}^{\text{3}}$.

Given information:

Assume length of a bathtub is $1.3\text{m}$, breadth of a bathtub is $0.5\text{m}$ and the height of a bathtub is $0.3\text{m}$. The density of copper is $8940\text{kg}/{\text{m}}^{\text{3}}$.

Formula to calculate the volume of the half portion of bathtub is,

$V=\frac{l\times b\times h}{2}$

• $l$ is the length of a bathtub.
• $b$ is the breadth of a bathtub.
• $h$ is the height of a bathtub.
• $V$ is the volume of volume of the half portion of bathtub.

Substitute $1.3\text{m}$ for $l$, $0.5\text{m}$ for $b$ and $0.3\text{m}$ for $h$ to find $V$.

$\begin{array}{c}V=1.3\text{m}\times 0.5\text{m}\times 0.3\text{m}\\ =0.195{\text{m}}^{\text{3}}\end{array}$

Section 2:

To determine: The order of magnitude of the mass of a bath tub half full of copper coins.

Answer: The order of magnitude of the mass of a bath tub half full of copper coins is $\sim {10}^3\text{kg}$.

Given information:

Assume length of a bathtub is $1.3\text{m}$, breadth of a bathtub is $0.5\text{m}$ and the height of a bathtub is $0.3\text{m}$. The density of copper is $8940\text{kg}/{\text{m}}^{\text{3}}$.

Formula to calculate the mass of half full of copper coins in the bathtub is,

$m_{\text{c}}={\rho }_{\text{c}}V$

• $m_{\text{c}}$ is the mass of the half full of copper coins in the bathtub.
• ${\rho }_{\text{c}}$ is the density of the copper.
• $V$ is the volume of a bathtub.

Substitute $8940\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ and $0.195{\text{m}}^{\text{3}}$ for $V$ to find $m$.

$\begin{array}{c}m=\left(8940\text{kg}/{\text{m}}^{\text{3}}\right)\left(0.195{\text{m}}^{\text{3}}\right)\\ =17433\text{kg}\end{array}$

Since, the mass of half full of copper coins is $17433\text{kg}$ and can be rewrite as $1.74\times {10}^3\text{kg}$. So, the order of magnitude is $\sim {10}^3\text{kg}$.

Conclusion:

Therefore, the order of magnitude of the mass of a bath tub half full of copper coins is $\sim {10}^3\text{kg}$.