An asteroid with a diameter of and a mass $\mathbf{10}\mathbf{Km}$of impacts the earth $\mathbf{2}\mathbf{.}\mathbf{60}\mathbf{\times }{\mathbf{10}}^{\mathbf{15}}\mathbf{}\mathbf{Kg}$at a speed of $\mathbf{32}\mathbf{.}\mathbf{0}\mathbf{Km}$Km landing in the Pacific Ocean. If $\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{\%}$ of the asteroid’s Kinetic energy goes to boiling ocean water (assuming an initial water temperature of $\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{°}\mathbf{}\mathbf{C}$),what mass of water will be boiled away by the collision?(For comparison , the mass of water contained in Lake Superior is about $\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{15}}\mathbf{Kg}\mathbf{}$. )

Defining Kinetic energy

It is the energy possessed by a body of massMwhich is moving the velocity V.

Formula of Kinetic energy

$K.E=\frac{1}{2}M{V}^2$

Where, K.E is the Kinetic energy of a body.

Defining the Latent Heat ofvaporization

The amount of heat required to change the liquid state of matter to gaseous state without raising its temperature is called Latent heat.

Formula of Latent Heat

Q = mL

Where, Q is the Latent heat, M is the mass of body and L is the specific heat of a substance.

Using formula:

$\mathrm{K}.\mathrm{E}=\frac{1}{2}{\mathrm{M}}_{\mathrm{asteroid}}{\left(\mathrm{V}\right)}^2$

Now, putting the values of constant in above equation

$$\begin{gathered} \mathrm{K}.\mathrm{E}=\frac{1}{2}\times 2.60\times {10}^{15}\mathrm{kg}\times {(32\times {10}^3)}^2{\mathrm{ms}}^{-1} \\ \mathrm{K}.\mathrm{E}=1.33\times {10}^{24}\mathrm{J} \end{gathered}$$

Thus, the kinetic energy of asteroid is $1.33\times {10}^{24}\mathrm{J}$

According to question $1.00\%$of the asteroid’s Kinetic energy goes to boiling ocean water.

So,

$$\begin{gathered} {\mathrm{Q}}_{\mathrm{w}}=0.01\times \mathrm{K}.\mathrm{E} \\ {\mathrm{M}}_{\mathrm{w}}{\mathrm{L}}_{\mathrm{vap}}=0.01\times 1.33\times {10}^{24}\mathrm{J} \\ {\mathrm{M}}_{\mathrm{W}}=\frac{0.01\times 1.33\times {10}^{24}\mathrm{J}}{2256\times {10}^3\mathrm{J}/\mathrm{kg}} \\ {\mathrm{M}}_{\mathrm{W}}=5.05\times {10}^{15}\mathrm{kg} \end{gathered}$$

Thus, the mass of water will be boiled away by the collision of asteroid is localid="1668167442438" $5.05\times {10}^{15}\mathrm{kg}$.