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## Q. 2.25

A system consists of a stone of mass 10 kg and a bucket containing 220 kg of water having same temperature. The stone falls into water from a height of 100 m. Calculate (a) the change in internal energy, (b) change in kinetic energy, (c) change in potential energy, (d) heat transferred, and (e) work transferred for the following cases:

(i) The instant the stone is about to enter water.
(ii) Just after the stone comes to rest in the bucket.
(iii) The temperature of stone and water remains the same.

## Verified Solution

(i) According to the first law of thermodynamics,

δQ – δW = dU + d(KE) + d(PE)

If δQ = 0 and δW = 0, then

dU + d(KE) + d(PE) = 0

For a non-flow process, dU = 0

$\therefore \quad$ d(KE) = – d(PE)

Or $\quad \Delta K E=-\Delta P E=-m g\left(z_2-z_1\right)$

= – 10 × 9.81(0 – 100) = 9810 J or 9.81 kJ

ΔPE = – 9.81 kJ

(ii)               δQ = 0, δW = 0 and d(KE) = 0

d(PE) = – dU

Or               ΔPE = – ΔU

$\Delta P E=m g\left(z_2-z_1\right)=10 \times 9.81(0-100)=-9.81 \ kJ$

(iii)               dU = 0, d(KE) = 0, and  δW = 0

$\therefore \quad$     $Q_{1-2}=\Delta P E=m g\left( z _2- z _1\right)=-9.81 \ kJ$