In air at 0 degree C, a 1.68-kg copper block at 0 degree C is set sliding at 2.3

Question

In air at 0 degree C, a 1.68-kg copper block at 0 degree C is set sliding at 2.30 m/s over a sheet of ice at 0 degree C. Friction brings the block to rest. Find the mass of the ice that melts. (Assume the latent heat of fusion for water is 3.33 Times 10^5 J/kg.) As the block slows down, identify its energy input Q, its change in internal energy Delta E_int, and the change in mechanical energy for the block-ice system. For the ice as a system, identify its energy input Q and its change in internal energy Delta E_int. A 1.68-kg block of ice at 0 degree C is set sliding at 2.30 m/s over a sheet of copper at 0 degree C. Friction brings the block to rest. Find the mass of the ice that melts. Evaluate Q and Delta E_int for the block of ice as a system and Delta E_mech for the block-ice system. Evaluate Q and Delta E_int for the metal sheet as a system.

Explanation / Answer

work done = change in kinetic energy

work done = initial kinetic energy - final kinetic energy

work done = 0.5 * 1.68 * 2.3^2 - 0

work done = 4.4436 J

heat = latent heat of fusion * mass

4.4436 = 3.33 * 10^5 * mass

mass = 0.00001334414 kg or 13.34414 mg

mass of ice that melts = 13.34414 mg

b) Q = -4.4436 J as block loses heat in order to melt ice

c) Q = 4.4436 J as ice gains heat

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