(a) In air at 0 degree C, a 1.50-kg copper block at 0 degree C is set sliding at

Question

(a) In air at 0 degree C, a 1.50-kg copper block at 0 degree C is set sliding at 2.20 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.) (b) As the block slows down, identify its energy input Q, its change in internal energy Delta and the change in mechanical energy for the block-ice system. (c) For the ice as a system, identify its energy input Q and its change in internal energy Delta E_int. (d) A 1.50-kg block of ice at 0 degree C is set sliding at 2.20 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. (e) Evaluate Q and Delta E_int for the block of ice as a system and Delta E_mech for the block-ice system. (f) Evaluate Q and Delta E_int for the metal sheet as a system. (g) A thin, 1.50-kg slab of copper at 16.0 degree C is set sliding at 2.20 m/s over an identical stationary slab at the same temperature, friction quickly stops the motion. Assuming no energy is transferred to the environment by heat, find the change in temperature of both objects. (Assume the specific heat of copper is 387 J/Kg middot degree C) (h) Evaluate Q and Delta E_int for the sliding slab and Delta E_mech for the two-slab system. (i) Evaluate Q and Delta E_int for the stationary slab.

Explanation / Answer

(a) friction energy =change in kinetic energy=0.5*m*v*v=3.63 J

We know friction =energy of ice melt

Let ice melted is xkg

X*333000=friction

So X=1.09*10(-5) kg or 0.0109 gram

(b)block is not shown but internal energy will increase

(c)for ice

Initial energy is more and internal energy of ice will increase

(d)same as part (a)

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