Chapter 9, Problem 87P

(a)

To determine

The scale readings in grams.

Answer to Problem 87P

The scale reading is $81.1\text{g}$.

Explanation of Solution

Dimensions of aluminum block is $2.00\text{cm}$ by $3.00\text{cm}$ by $5.00\text{cm}$.

The spring scale shows the mass of block.

Write the equation to find the mass of block.

$m={\rho }_{al}V$

Here, $m$ is the mass of block, $\rho$ is the density of aluminum, and $V$ is the volume of block.

Conclusion:

Substitute $2702\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{al}$ and $2.00\text{cm}\times 3.00\text{cm}\times 5.00\text{cm}$ for $V$ in the above equation to find $m$.

$\begin{array}{c}m=\left(2702\text{kg}/{\text{m}}^{\text{3}}\right)\left(\left(2.00\text{cm}\times 3.00\text{cm}\times 5.00\text{cm}\right)\left(\frac{{10}^{-6}{\text{m}}^{\text{3}}}{1{\text{cm}}^{\text{3}}}\right)\right)\\ =8.11\times {10}^{-2}\text{kg}\left(\frac{{\text{10}}^{\text{3}}\text{g}}{1\text{kg}}\right)\\ =81.1\text{g}\end{array}$

Therefore, the scale reading is $81.1\text{g}$.

(b)

To determine

The readings of two scales.

Answer to Problem 87P

Readings are $55.6\text{g}$ and $486\text{g}$ respectively.

Explanation of Solution

Density of oil is $850\text{kg}/{\text{m}}^{\text{3}}$ and reading of the second scale is $460.0\text{g}$.

Write the equation to find the reading of first scale.

$m_1=m-{\rho }_{oil}V$ (I)

Here, $m_1$ is the mass of first scale and ${\rho }_{oil}$ is the density of oil.

Write the equation to find the scale reading with block immersed in the oil.

$m_2=m_b+{\rho }_{oil}V$ (II)

Here, $m_2$ is the scale reading with block immersed in the oil and $m_b$ is the mass of beaker with oil.

Conclusion:

Substitute $81.1\text{g}$ for $m$, $850\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{oil}$, and $2.00\text{cm}\times 3.00\text{cm}\times 5.00\text{cm}$ for $V$ in equation (I) to find $m_1$.

$\begin{array}{c}{m}_1=81.1\text{g}-\left(\left(850\text{kg}/{\text{m}}^{\text{3}}\left(\frac{{\text{10}}^{\text{3}}\text{g}}{\text{1}\text{kg}}\right)\right)\left(\left(2.00\text{cm}\times 3.00\text{cm}\times 5.00\text{cm}\right)\left(\frac{{10}^{-6}{\text{m}}^{\text{3}}}{1{\text{cm}}^{\text{3}}}\right)\right)\right)\\ =81.1\text{g}-25.5\text{g}\\ \text{=55}\text{.6}\text{g}\end{array}$

Substitute $460.0\text{g}$ for $m_b$, $850\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{oil}$, and $2.00\text{cm}\times 3.00\text{cm}\times 5.00\text{cm}$ for $V$ in equation (II) to find $m_2$.

$\begin{array}{c}{m}_2=460.0\text{g}+\left(\left(850\text{kg}/{\text{m}}^{\text{3}}\right)\left(\left(2.00\text{cm}\times 3.00\text{cm}\times 5.00\text{cm}\right)\left(\frac{{10}^{-6}{\text{m}}^{\text{3}}}{1{\text{cm}}^{\text{3}}}\right)\right)\right)\\ =460.0\text{g}+26.0\text{g}\\ \text{=486}\text{.0}\text{g}\end{array}$

Therefore, the readings are $55.6\text{g}$ and $486\text{g}$ respectively.