An Iron block )m=1kg) initially held stationaryis dropped from a 100 m height. I

Question

An Iron block )m=1kg) initially held stationaryis dropped from a 100 m height. It strikes the ground and is assumed to be perfectly insulated from its environment. Calculate the temperature change in the block after it comes to rest on the ground. Assuming Cp of iron=c = 0.447 kJ/kg K.

A. Calculate the temperature increase of the block if it has a c=0.447kj/kg*k

B. What is the kinetic energy at impact?

Knowing that the lost work in a given process is proportional to the entropy generated in that process by the formula W=t*sigma, where T is the reservior temperature ultimately receiving the process' generated entropy.

C. Calculate the entropy generated as the block comes to rest adiabaticaly.

D. What assumptions must be made reguarding the process prior to doing the calculations above?

Explanation / Answer

(A) 1/2 mv^2 = mc Delta T => Delta T = 1 /0.447 = 2.24 Kelvin

(B) 1/2 mv^2 = mgh = 1 kJ

(C) entropy increase of surrounding = (1 kJ) / T

(D) Assumptions:
1. Neglect the air resistance.
2. While falling the block is in the equilibrium all the time.
3. The collision is completely inelastic.

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