When falling from what height, the water could boil on impact on the ground, if half of the mechanical

When falling from what height, the water could boil on impact on the ground, if half of the mechanical energy of the water was converted into its internal energy? The initial temperature of the water is 20 degrees

Given:

T0 = ​​20 degrees Celsius – initial water temperature;

T1 = 100 degrees Celsius – boiling point of water;

c = 4183 J / kg * g – specific heat capacity of water;

g = 10 m / s ^ 2 – acceleration of gravity.

It is required to determine the height H, upon falling from which the water would boil if half of the mechanical energy would be converted into internal energy.

The energy at which the water will boil is:

Q = c * m * (T – T0).

This energy should be equal to half the mechanical energy, that is:

Q = m * H * g / 2.

m * H * g / 2 = c * m * (T – T0).

H * g / 2 = c * (T – T0).

H = 2 * c * (T – T0) / g = 2 * 4183 * (100 – 20) / 10 = 2 * 4183 * 80/10 = 66928 meters.

Answer: water will boil when dropped from a height of 66,928 meters (67 kilometers).