physics problem (a) In air at 0 Degree C, a 1.58-kg copper block at 0 Degree C i

Question

physics problem

(a) In air at 0 Degree C, a 1.58-kg copper block at 0 Degree C is set sliding at 2.95 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.) ____________ mg (b) 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. Q = _________ J Delta E_int = __________ J Delta E_mech = ___________ J (c) For the ice as a system, identify its energy input Q and its change in internal energy Delta E_int. Q = ___________ J Delta E_int = ___________ J (d) A 1.58-kg block of ice at 0 Degree C is set sliding at 2.95 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. __________ mg (e) Evaluate Q and Delta E_int for the block of ice as a system and Delta E_mech for the block-ice system. Q = ____________ J Delta E_int = ___________ J Delta E_mech = ____________ J (f) Evaluate Q and Delta E_int for the metal sheet as a system. Q = __________ J Delta E_int = ____________ J (g) A thin, 1.58-kg slab of copper at 25.0 Degree C is set sliding at 2.95 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.) Delta T_sliding slab = ____________ Degree C Delta T_stationary slab = ______________ Degree C (h) Evaluate Q and Delta E_int for the sliding slab and Delta E_mech for the two-slab system. Q = ___________ J Delta E_int = ______________ J Delta E_mech = ____________ J (i) Evaluate Q and Delta E_int for the stationary slab. Q = _____________ J Delta E_int = ____________ J

Explanation / Answer

a. mass of copper block, m = 1.58 kg

temperature, T = 0 C

velocity, v = 2.95 m/s

its sliding over a sheet of ice

KE = 0.5mv^2

this eneregy is lost as friction brings the block to stop

this energy is transferred to ice which melts

so mass of ice that mel;ts be M

then ML = 0.5mv^2

where L = 3.33*10^5 J/kg

so, M = 20.6455 micro grams

b. block ice system

Q = energy input = 0 J

dEint = 6.874975 J

dEmech = -6.874975 J

c. for ice

Q = 6.874975 J

dEint = 6.874975 J

d. the same amountr of ice will melt as the same amount of KE is there in theblock ( same mass, smae velocity)

so, M = 20.6455 micro grams

e. For ice

Q = energy input = 6.874975 J

dEint = 6.874975 J

dEmech = -6.874975 J

f. for metal sheet

Q = 0 J

dEint = 0 J

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