Chapter 14, Problem 128QAP

To determine

(a)

The maximum amount of ocean water that could have been evaporated.

Answer to Problem 128QAP

The maximum amount of ocean water that could have been evaporated is $2.48\times {10}^{16}\text{kg}$.

Explanation of Solution

Given info:

Diameter of the spherical asteroid is, $D=10.0\text{km}$.

Diameter of the spherical asteroid is, $\rho =2.0\text{g}/{\text{cm}}^3$.

Velocity of the spherical asteroid is, $v=11\text{km}/\text{s}$.

Initial temperature of the ocean water is, $T_0=20.0°\text{C}$.

Formula used:

Formula for the energy balance equation is,

  $\text{Kinetic}\text{energy}=\text{Heat}\text{absorbed}\text{by}\text{water}\text{to}\text{evaporate}$

Calculation:

The temperature difference in Kelvin is calculated as,

  $\begin{array}{c}\Delta T=T_f-T_0\\ \Delta T=\left(100+273\right)°\text{K}-\left(20+273\right)°\text{K}\\ \Delta T=80\text{K}\end{array}$

The mass of the asteroid can be calculated as,

  $\begin{array}{c}{m}_{\text{asteroid}}=\rho \times \left(\frac{4}{3}\pi {\left(\frac{D}{2}\right)}^3\right)\\ m_{\text{asteroid}}=2.0\times \frac{{\text{10}}^{-3}\text{kg}}{{\left({10}^{-2}\text{m}\right)}^3}\times \left(\frac{4}{3}\pi {\left(\frac{10.0\times {10}^3\text{m}}{2}\right)}^3\right)\\ m_{\text{asteroid}}=\left(2.0\times \frac{{\text{10}}^{-3}}{{10}^{-6}}\text{kg}/{\text{m}}^3\right)\times \left(\frac{4}{3}\pi {\left(5\times {10}^3\text{m}\right)}^3\right)\\ m_{\text{asteroid}}=1.05\times {10}^{15}\text{kg}\end{array}$

The total heat need by the water is equal to the sum total of heat required to raise temperature from $20°\text{C}$to the boiling point and the latent heat of vaporization. The total heat needed for the process is provided by the kinetic energy of the asteroid.

The energy balance equation becomes,

  $\begin{array}{c}\frac{1}{2}{m}_{\text{asteroid}}{v}^2=m_{\text{water}}c\Delta T+m_{\text{water}}{L}_{\text{v}}\\ \frac{1}{2}{m}_{\text{asteroid}}{v}^2=m_{\text{water}}\left(c\Delta T+L_{\text{v}}\right)\\ m_{\text{water}}=\frac{1}{2}\frac{m_{\text{asteroid}}{v}^2}{\left(c\Delta T+L_{\text{v}}\right)}\end{array}$

Substituting the values in the above equation, we get

  $\begin{array}{c}{m}_{\text{water}}=\frac{1}{2}\frac{\left(1.05\times {10}^{15}\text{kg}\right){\left(11\times 1000\text{m}/\text{s}\right)}^2}{\left(\left(4186\text{J}/\text{kg}\cdot \text{K}\right)\times 80\text{K}+2260\times {10}^3\text{J}/\text{K}\right)}\\ =2.48\times {10}^{16}\text{kg}\end{array}$

Conclusion:

Thus, the maximum amount of ocean water than could have been evaporated is $2.48\times {10}^{16}\text{kg}$.

To determine

(b)

The height of the water cube.

Answer to Problem 128QAP

The height of the water cube is $2.92\times {10}^4\text{m}$.

Explanation of Solution

Given info:

Mass of the water is, $m_{\text{water}}=2.48\times {10}^{16}\text{kg}$.

Formula used:

Formula for the volume of water cube is,

  $V=\frac{m}{\rho }$

Calculation:

The side of the water cube can be calculated as,

  $\begin{array}{c}V=\frac{m}{\rho }\\ L=\sqrt[3]{\frac{m}{\rho }}\end{array}$

Substituting the given values in the above equation, we get

  $\begin{array}{c}L=\sqrt[3]{\frac{2.48\times {10}^{16}\text{kg}}{1000\text{kg}/{\text{m}}^3}}\\ L=2.92\times {10}^4\text{m}\end{array}$

Conclusion:

Thus, the height of the water cube is $2.92\times {10}^4\text{m}$.