Question 14.1: EXPLOSION OVER AN ICE SHEET GOAL Calculate time of travel fo...

Calculate the speed of sound in air at -7.00^{\circ} \mathrm{C}, which is equivalent to 266 \mathrm{~K} :

v_{\text {air }} =(331 \mathrm{~m} / \mathrm{s}) \sqrt{\frac{T}{273 \mathrm{~K}}}=(331 \mathrm{~m} / \mathrm{s}) \sqrt{\frac{266 \mathrm{~K}}{273 \mathrm{~K}}}=327 \mathrm{~m} / \mathrm{s}

 Calculate the travel time through the air:

t_{\text {air }} =\frac{d}{v_{\text {air }}}=\frac{275 \mathrm{~m}}{327 \mathrm{~m} / \mathrm{s}}=0.841 \mathrm{~s}

Compute the speed of sound in ice, using the bulk modulus and density of ice:

v_{\text {ice }} =\sqrt{\frac{B}{\rho}}=\sqrt{\frac{9.2 \times 10^{9} \mathrm{~Pa}}{917 \mathrm{~kg} / \mathrm{m}^{3}}}=3.2 \times 10^{3} \mathrm{~m} / \mathrm{s}

Compute the travel time through the ice:

t_{\text {ice }} =\frac{d}{v_{\text {ice }}}=\frac{867 \mathrm{~m}}{3  200 \mathrm{~m} / \mathrm{s}}=0.27 \mathrm{~s}

Compute the travel time through the ocean water:

t_{\text {water }} =\frac{d}{v_{\text {water }}}=\frac{1  250 \mathrm{~m}}{1  533 \mathrm{~m} / \mathrm{s}}=0.815 \mathrm{~s}

Sum the three times to obtain the total time of propagation:

\begin{aligned}t_{\text {tot }} &=t_{\text {air }}+t_{\text {ice }}+t_{\text {water }}=0.841 \mathrm{~s}+0.27 \mathrm{~s}+0.815 \mathrm{~s} \\&=1.93 \mathrm{~s}\end{aligned}

REMARKS Notice that the speed of sound is highest in solid ice, second highest in liquid water, and slowest in air. The speed of sound depends on temperature, so the answer would have to be modified if the actual temperatures of ice and the sea water were known. At 0^{\circ} \mathrm{C}, for example, the speed of sound in sea water falls to 1449 \mathrm{~m} / \mathrm{s}.