A stone of 20 kg of mass is held 15 m above a tank containing 200 kg of water. T

Question

A stone of 20 kg of mass is held 15 m above a tank containing 200 kg of water. The stone, water and the air are at the same temperature initially. The stone then falls into the water. Choosing the stone as the system, and starting from a statement of the first law of thermodynamics, determine, change in U, change in PE, change in KE, Q and W, (magnitude and sign), when each of the following states is reached:
a. the stone is about to enter the water,
b. the stone has come to rest in the tank, (assume it decelerates instantly, as soon as it hits the water), and
c. when the stone and water return to their initial temperature.

Explanation / Answer

From the first law of thermodynamics Q = dU + dKE + dPE + W (a) Here heat transfer,Q = 0 Workdone,W = 0 Change in internal energy,dU = 0 But only potential energy exists due to elevation of body dPE = mg(z2-z1) = 20*9.81*(0-15) = -2943 J 0 = 0 + dKE + -2943 J + 0 dKE = 2943 J As the body reaches the surface of water, all the initial energy that the body possess as potential energy gets converted into kinetic energy (b) Here heat transfer,Q = 0 Workdone,W = 0 Change in potential energy,dPE = 0 But kinetic energy, dKE = 2943 J Hence 0 = dU + 2943 J + 0 + 0 dU = 2943 J As the body is left with kinetic energy as it reaches the top surface of water, only this kinetic energy is available energy and gets converted into internal energy (c) Workdone,W = 0 Change in potential energy,dPE = 0 kinetic energy, dKE = 0 Hence Q = dU + 0 + 0 + 0 Q = -2943 J As the body is left with internal energy as it reaches the bottom surface of water, only this kinetic energy is available energy and gets converted into heat energy

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