Solved: If you drop a piece of ice on a hard surface, the

Chapter 9, Problem 42TAS

(choose chapter or problem)

The quantity of heat with temperature change is \(Q=c m \Delta T\). For a change of phase of water it is Q = mL, where \(L_{f}\) is the heat of fusion, 80 cal/g, and \(L_{v}\) is the heat of vaporization, 540 cal/g.

If you drop a piece of ice on a hard surface, the energy of impact will melt some of the ice. The farther the ice drops, the more ice will melt upon impact. Show that to completely melt a block of ice that falls without air drag, it should ideally be dropped from a height of 34 km. [Hint: Equate the joules of gravitational potential energy to the product of the mass of ice and its heat of fusion (in SI units, 335,000 J/kg). Do you see why the answer doesn’t depend on mass?]